Unit 6 – Chapter 10

Unit 6 – Chapter 10

Unit 6• Chapter 9 Review and Chap. 9 Skills

• Section 10.1 – Graph y = ax2 + c

• Section 10.2 – Graph y = ax2 + bx +c

• Section 10.3 – Solve Quadratic Eqns by Graphing

• Section 10.4 – Solve Quad. Eqns. w/Square roots

• Section 10.6 – Solve Quad. Eqns. With Quad. Formula

• Section 10.7-10.8 – Compare Linear, Exponential, and Quadratic Models

Warm-Up – Chapter 10

Prerequisite Skills VOCABULARY CHECK

exponential functionANSWER

1. The x-coordinate of a point where a graph crosses the x-axis is a (n) .?

ANSWER x-intercept

2. A (n) is a function of the form y = a bx where a 0,b > 0, and b 1. = =

?

Copy and complete the statement.

Prerequisite Skills SKILLS CHECK

Draw the blue figure. Then draw its image after a reflection in the red line.

3.

ANSWER


Draw the blue figure. Then draw its image after a reflection in the red line.

4.

ANSWER


Graph y1 = x and y2 = x – 5. Describe the similarities and differences in the graphs.

ANSWER

•Graphs are Parallel•Have the sameSlope•Y2 is translatedDown 5 units

Lesson 10.1, For use with pages 628-634

1. Graph the function y = 2x.

2. Identify the domain and range of your graph inExercise 1.

ANSWER domain: all real numbers; range: all positive real numbers

ANSWER


1. Graph the function y = x2.



ANSWER


1. Graph the function y = 3x2.



ANSWER


1. Graph the function y = (0.1)x2.



ANSWER


1. Graph the function y = x2 + 2


ANSWER domain: all real numbers; range: all positive real numbers > 2

ANSWER


1. Graph the function y = (-1)x2


ANSWER domain: all real numbers; range: all negative numbers

ANSWER

Vocabulary – 10.1• Quadratic Function

• Non-linear function in the form y = ax2 + bx + c where a ≠ 0

• Parabola

• U – shaped graph of a quadratic function

• Parent quadratic function

• Y = x2

• Vertex of a Parabola

• Lowest (or highest) point on a parabola

• Axis of symmetry

• Line that passes through the vertex

• Divides the parabola into two symmetric parts

Notes – 10.1 – Graph y = ax2 + c• All Quadratic Function graphs look like what?

•A parabola•What happens to the shape of the parabola if a >1?

•It is a vertical stretch•It gets skinny!

•What happens to the shape of a parabola if a < 1?•Vertical shrink•It gets fat (or “fluffy”!)!

•What happens if a < 0?•The graph curves down.

•What effect does “c” have on the graph of the function?•It translates it up or down.

Examples 10.1

EXAMPLE 1 Graph y= ax2 where a > 1

STEP 1

Make a table of values for y = 3x2

x – 2 – 1 0 1 2

y 12 3 0 3 12

Plot the points from the table.

STEP 2

EXAMPLE 1

STEP 3

Draw a smooth curve through the points.

Compare the graphs of y = 3x2 and y = x2. Both graphs open up and have the same vertex, (0, 0), and axis of symmetry, x = 0. The graph of y = 3x2 is narrower than the graph of y = x2 because the graph of y = 3x2 is a vertical stretch (by a factor of 3) of the graph of y = x2.

STEP 4

Graph y= ax2 where a > 1

EXAMPLE 2 Graph y = ax2 where a < 1

Graph y = 14

– x2. Compare the graph with the graph ofy = x2.

STEP 1

Make a table of values for y =14

– x2.

x – 4 – 2 0 2 4

y – 4 – 1 0 – 1 – 4

EXAMPLE 2

STEP 2



STEP 3

Graph y = ax2 where a < 1

EXAMPLE 2

STEP 4

Compare the graphs of y =14

– x2. and y = x2.

Both graphs have the same vertex (0, 0), and the same axis of symmetry, x = 0. However, the graph of 1

4– x2.y =

is wider than the graph of y = x2 and it opens down. This is because the graph of 1

4– x2.y = is a vertical shrink

by a factor of14 with a refl ection in the x-axis of the

graph of y = x2.

Graph y = ax2 where a < 1

EXAMPLE 3 Graph y = x2 + c

Graph y = x2 + 5. Compare the graph with the graph of y = x2.

STEP 1

Make a table of values for y = x2 + 5.

x – 2 – 1 0 1 2

y 9 6 5 6 9


STEP 2


STEP 3



STEP 4

Compare the graphs of y = x2 + 5 and y = x2. Both graphs open up and have the same axis of symmetry, x = 0. However, the vertex of the graph of y = x2 + 5, (0, 5), is different than the vertex of the graph of y = x2, (0, 0), because the graph of y = x2 + 5 is a vertical translation (of 5 units up) of the graph of y = x2.

GUIDED PRACTICE for Examples 1, 2 and 3

Graph the function. Compare the graph with the graph of x2.

1. y= –4x2

STEP 1

Make a table of values for y = –4x2

x – 2 – 1 0 1 2

Y – 16 – 4 0 – 4 – 16


STEP 2



STEP 3


STEP 1

x – 6 – 3 0 3 6

y 12 3 0 3 12

2. y = x213

Make a table of values for y = x213


STEP 2



STEP 3


3. y = x2 +2

STEP 1

Make a table of values for y = x2 +5

x – 2 – 1 0 1 2

Y 6 3 2 3 6


STEP 2


STEP 3


EXAMPLE 4

Graph y = x2 – 4. Compare the graph with the graph of y = x2.

12

STEP 1

Make a table of values for y = x2 – 4.12

x – 4 – 2 0 2 4

y 4 – 2 – 4 –2 4

Graph y = ax2 + c

EXAMPLE 4 Graph y = ax2 + c

STEP 2


STEP 3


EXAMPLE 4 Graph y = ax2 + c

STEP 4

Compare the graphs of y = x2 – 4 and y = x2. Both

graphs open up and have the same axis of symmetry,

x = 0. However, the graph of y = x2 – 4 is wider and

has a lower vertex than the graph of y = x2 because the

graph of y = x2 is a vertical shrink and a vertical

translation of the graph of y = x2.

12

12

12

GUIDED PRACTICE for Example 4

Graph the function. Compare the graph with the graph of x2.

4. y= 3x2 – 6

STEP 1

Make a table of values for y = 3x2 – 6.

x – 2 – 1 0 2 1

y 6 – 3 – 6 6 – 3


STEP 2


STEP 3



5. y= –5x2 + 1

STEP 1

Make a table of values for y = –5x2 + 1.

x – 2 – 1 0 1 2

y –19 – 4 1 –4 – 19


STEP 2


STEP 3



6. y = x2 – 2.34

STEP 1

Make a table of values for y = x2 – 2.34

x – 4 – 2 0 4 2

y 10 1 – 2 10 1


STEP 2


STEP 3


EXAMPLE 5 Standardized Test Practice

How would the graph of the function y = x2 + 6 be affected if the function were changed to y = x2 + 2?

A The graph would shift 2 units up.

B The graph would shift 4 units up.

C The graph would shift 4 units down.

D The graph would shift 4 units to the left.


SOLUTION

The vertex of the graph of y = x2 + 6 is 6 units above the origin, or (0, 6). The vertex of the graph of y = x2 + 2 is 2 units above the origin, or (0, 2). Moving the vertex from (0, 6) to (0, 2) translates the graph 4 units down.

ANSWER

The correct answer is C. A B C D

EXAMPLE 6 Use a graph

SOLAR ENERGY

A solar trough has a reflective parabolic surface that is used to collect solar energy. The sun’s rays are reflected from the surface toward a pipe that carries water. The heated water produces steam that is used to produce electricity.

The graph of the function y = 0.09x2 models the cross section of the reflective surface where x and y are measured in meters. Use the graph to find the domain and range of the function in this situation.

EXAMPLE 6 Use a graph

SOLUTION

STEP 1

Find the domain. In the graph, the reflective surface extends 5 meters on either side of the origin. So, the domain is 5 ≤ x ≤ 5.

STEP 2Find the range using the fact that the lowest point on the reflective surface is (0, 0) and the highest point, 5, occurs at each end.

y = 0.09(5)2 = 2.25 Substitute 5 for x. Then simplify.

The range is 0 ≤ y ≤ 2.25.

GUIDED PRACTICE for Examples 5 and 6

Describe how the graph of the function y = x2+2 would be affected if the function were changed to y = x2 – 2.

7.

SOLUTION

The vertex of the graph of y = x2 + 2 is 6 units above the origin, or (0, 2). The vertex of the graph of y = x2 – 2 is 2 units down above the origin, or (0, – 2). Moving the vertex from (0, 2) to (0, – 2) translates the graph 4 units down.

ANSWER

The graph would be translated 4 unit down.


What if In example 6 ,suppose the reflective surface extends just 4 meters on either side of the origin.Find the range of the function in this situation.

8.

SOLUTION

STEP 1

Find the domain. In the graph, the reflective surface extends 4 meters on either side of the origin. So, the domain is – 4 ≤ x ≤ 4.


STEP 2Find the range using the fact that the lowest point on the reflective surface is (0, 0) and the highest point 4, occurs at each end.

y = 0.09(4)2 = 144 Substitute 5 for x. Then simplify.

The range is – 4 ≤ x ≤ 4, 0 ≤y ≤1.44.

Warm-Up – 10.2

Evaluate the expression.

1. x2 – 2 when x = 3

2. 2x2 + 9 when x = 2

ANSWER 7

ANSWER 17


ANSWER 6.25 in.2



Martin is replacing a square patch of counter top. The area of the patch is represented by A = s2. What is the area of the patch if the side length is 2.5 inches?

3.

Vocabulary –10.2• Maximum Value

• Highest possible value of a graph

• Not all graphs have a maximum value

• Minimum Value

• Smallest possible value of a graph

• Not all graphs have a minimum value

• Vertex

• AKA the “maximum” or “minimum” value of a parabola


Graph the following equation.

1) Graph y = 3x2 - 12x + 5.2) Press 2nd Calc Minimum3) Pick two points, one to the left and one to the right of the minimum.4) What is the X coordinate of that point??

ANSWER X-coordinate is 2

1) Using the equation y = 3x2 - 12x + 5, calculate the following on your white board.

2) x = - b / (2a)3) What do you get? Notice anything????

ANSWER X = 2


Graph the following equation. 1) What seems to be true about the x coordinate of the vertex (minimum in this case) and the equation

x = - b / (2a)?

ANSWERThey appear to be the same!

1) I bet it only works once!!!


Graph the following equation. 1) What is the line of symmetry for this parabola?

2) Use the Table function to find the y-intercept (Where x = 0).

1) I bet it only works once!!!

ANSWER1) x = 22) y-intercept (0,5)

Notes – 10.2 – Graph ax2 + bx + c• PROPERTIES OF GRAPH OF y = ax2 + bx + c

1. If a > 0 opens up and opens down if a < 02. If |a| > 1, gets “skinnier” & if |a| < 1, “fluffier”3. Vertex has an x-coordinate of – b / (2a)4. Line of symmetry is x = – b / (2a)5. Y-intercept of (0,c)

•TO FIND MINIMUM/MAXIMUM POINTS1. Find the x-coordinate of the vertex and plug it

in to the equation y = ax2 + bx + c

Examples 10.2

EXAMPLE 1 Find the axis of symmetry and the vertex

Consider the function y = – 2x2 + 12x – 7.

a. Find the axis of symmetry of the graph of the function.

b. Find the vertex of the graph of the function.

EXAMPLE 1 Find the axis of symmetry and the vertex

SOLUTION

122(– 2)x = – b

2a= = 3 Substitute – 2 for a and 12 for b.

Then simplify.

For the function y = –2x2 + 12x – 7, a = 2 and b = 12.a.

b. The x-coordinate of the vertex is , or 3. b 2a

–

y = –2(3)2 + 12(3) – 7 = 11 Substitute 3 for x. Then simplify.

ANSWER The vertex is (3, 11).

EXAMPLE 2 Graph y = ax2 + b x + c

Graph y = 3x2 – 6x + 2.

Determine whether the parabola opens up or down. Because a > 0, the parabola opens up.

STEP 1

STEP 2 =Find and draw the axis of symmetry: x = – b

2a– – 62(3)= 1.

STEP 3Find and plot the vertex.

EXAMPLE 2

The x-coordinate of the vertex is – b2a

, or 1.

To find the y-coordinate, substitute 1 for x in the function and simplify.

y = 3(1)2 – 6(1) + 2 = – 1

So, the vertex is (1, – 1).

STEP 4Plot two points. Choose two x-values less than the x-coordinate of the vertex. Then find the corresponding y-values.

Graph y = ax2 + b x + c


x 0 – 1

y 2 11

STEP 5Reflect the points plotted in Step 4 in the axis of symmetry.

STEP 6Draw a parabola through the plotted points.

EXAMPLE 2 Graph y = ax2 + b x + c


1. Find the axis of symmetry and vertex of the graph of the function.

y = x2 – 2x – 3.

SOLUTION

Substitute – 2 for b and 1 for a. Then simplify.

For the function y = x2 – 2x – 3, a = 1 and b = – 2.

The x-coordinate of the vertex is , or 1. b 2a

–

–x = – b 2a

= = 1 – 22 1


To find the y-coordinate, substitute 1 for x in the function and simplify.

y = (1)2 – 2(1) – 3 = – 4

The vertex is (1, – 4).

Substitute 1 for x. Then simplify.


2. Graph the Function. y = 3x2 + 12x – 1. Label the vertex and axis of symmetry.

Determine whether the parabola opens up or down. Because a > 0, the parabola opens up.

STEP 1

STEP 2Find and draw the axis of symmetry: =x = – b

2a––122(3)

= – 2.

STEP 3Find and plot the vertex.


The x-coordinate of the vertex is – b2a

, or – 2.

To find the y-coordinate, substitute –2 for x in the function and simplify.

y = 3(– 2)2 + 12(– 2) – 1 = – 14

So, the vertex is (– 2, – 14).


x 0 – 1

y – 1 – 10

STEP 4

Plot two points. Choose two x-values less than the x-coordinate of the vertex. Then find the corresponding y-values.


STEP 5

Reflect the points plotted in Step 4 in the axis of symmetry.

STEP 6

Draw a parabola through the plotted points.

EXAMPLE 3 Find the minimum or maximum value

Tell whether the function f(x) = – 3x2 – 12x + 10 has aminimum value or a maximum value. Then find theminimum or maximum value.

SOLUTION

Because a = – 3 and – 3 < 0, the parabola opens down andthe function has a maximum value. To find the maximumvalue, find the vertex.

x = – = – = – 2 b2a

– 122(– 3)

f(– 2) = – 3(– 2)2 – 12(– 2) + 10 = 22 Substitute 22 for x. Thensimplify.

The x-coordinate is – b2a

Find the minimum or maximum valueEXAMPLE 3

ANSWER

The maximum value of the function is f(– 2) = 22.

Find the minimum value of a functionEXAMPLE 4

The suspension cables between the two towers of the Mackinac Bridge in Michigan form a parabola that can be modeled by the graph of y = 0.000097x2 – 0.37x + 549 where x and y are measured in feet.What is the height of the cable above the water at its lowest point?

Suspension Bridges

Find the minimum value of a functionEXAMPLE 4

SOLUTION

The lowest point of the cable is at the vertex of theparabola. Find the x-coordinate of the vertex. Use a = 0.000097 and b = – 0.37.

x = – = – = 1910 b2a

– 0.372(0.000097)

Use a calculator.

Substitute 1910 for x in the equation to find they-coordinate of the vertex.

≈ 0.000097(1910)2 – 0.37(1910) + 549 ≈ 196

EXAMPLE 4

ANSWER

The cable is about 196 feet above the water at its lowest point.

Find the minimum value of a function


3. Tell whether the function f(x) = 6x2 + 18x + 13 has aminimum value or a maximum value. Then find theminimum or maximum value.

SOLUTION

Because a = 6 and 6 > 0, the parabola opens down andthe function has a minimum value. To find the minimumvalue, find the vertex.

1 2

The minimum value is


Suspension Bridges

4. The cables between the two towers of the Takoma Narrow Bridge form a parabola that can be modeled by the graph of the equation y = 0.00014x2 – 0.4x + 507 where x and y are measured in feet.What is the height of the cable above the water at its lowest point? Round your answer to the nearest foot.


SOLUTION

The lowest point of the cable is at the vertex of theparabola. Find the x-coordinate. Use a = 0.00014 and b = – 0.4.

x = – = – = 1428 b2a

– 0.42(0.00014)

Use a calculator.

Substitute 1428 for x in the equation to find they-coordinate of the vertex.

y = 0.00014(1428)2 – 0.4(1428) + 507 = 221

ANSWER

The cable is about 221 feet above the water at its lowest point.


Warm-Up – 10.3 – Solve Quad. Eqns by Graphing

Find the zeros of the polynomial function BY FACTORING!

1. f(x) = x2 + 5x – 36

2. f(x) = x2 – 9x + 20

ANSWER –9, 4

ANSWER 4, 5


Find the line of symmetry and the vertex of the following:

1. f(x) = x2 + 5x – 36

2. f(x) = x2 – 9x + 20

ANSWER Line of symmetry x = -2.5Vertex = (-2.5,-42.3)

ANSWER Line of symmetry x = 4.5Vertex = (4.5,-0.25)


Find the line of symmetry and the minimum of the following:

ANSWER 9 in.

Solve.

You are making a rectangular flag for your school team. The number of square inches in the area of the flag is 9x + 108. The number of inches in the length of the flag is x + 12. What is the width of the flag?HINT: DRAW A PICTURE ON YOUR WHITEBOARD!

3.


Find the zeros of the polynomial function by graphing.

1. f(x) = x2 + 5x – 36

2. f(x) = x2 – 9x + 20

ANSWER –9, 4

ANSWER 4, 5


BONUS: Find the average of the x-coordinates of the roots. Look familiar? Bet it only works once!

BONUS: Find the average of the x-coordinates of the roots. Look familiar? Did it work again?!??

Vocabulary – 10.3 – • Quadratic Equation

• An equation that can be written in the form

f(x) = ax2 + bx + c

• roots

• Where an equation crosses the X-axis.

• AKA the ZEROS of a function.

• Found by setting an equation = 0 and solving for x.

Notes – 10.3–Solve Quad. Eqns by Graphing.• FINDING ZEROS OF A QUAD. FUNCTION

1. Factoring2. Checking a table3. Graphing and finding the “zero”

•Quadratic Equations can have a MAXIMUM of _____ roots, but it can have _____ or even _______.•Most of the time, the zeros of a function are NOT integers!

2 1

NONE

Examples 10.3

EXAMPLE 1 Solve a quadratic equation having two solutions

Solve x2 – 2x = 3 by graphing.

STEP 1Write the equation in standard form.

Write original equation.x2 – 2x = 3Subtract 3 from each side.x2 – 2x – 3 = 0

STEP 2

Graph the function y = x2 – 2x – 3. The x-intercepts are – 1 and 3.

SOLUTION

EXAMPLE 1

ANSWER

The solutions of the equation x2 – 2x = 3 are – 1 and 3.

Solve a quadratic equation having two solutions

You can check – 1 and 3 in the original equation.

x2 – 2x = 3 x2 – 2x = 3

(–1)2 –2(–1) 3=? (3)2 –2(3) 3=

?

3 = 3 3 = 3

Write original equation.

Substitute for x.

Simplify. Each solution checks.

CHECK:

EXAMPLE 2 Solve a quadratic equation having one solution

Solve – x2 + 2x = 1 by graphing.

SOLUTION


Write original equation.– x2 + 2x = 1

Subtract 1 from each side.– x2 + 2x – 1 = 0

EXAMPLE 2 Solve a quadratic equation having one solution

STEP 2

Graph the function y = – x2 + 2x – 1.

The x-intercept is 1.

ANSWER The solution of the equation – x2 + 2x = 1 is 1.

EXAMPLE 3 Solve a quadratic equation having no solution

Solve x2 + 7 = 4x by graphing.

SOLUTION


Write original equation.x2 + 7 = 4x

Subtract 4x from each side. x2 – 4x + 7 = 0

EXAMPLE 3 Solve a quadratic equation having no solution

STEP 2Graph the function y = x2 – 4x + 7.

ANSWER The equation x2 + 7 = 4x has no solution.


Solve the equation by graphing.

1. x2 – 6x + 8 = 0

Graph the function y = x2 – 6x + 8

The x-intercept 2 and 4.

ANSWER

The solution of the equation x2 – 6x + 8 are 2 and 4.


CHECK:

You can check – 1 and 3 in the original equation.

x2 – 6x + 8 = 0

(2)2 – 6 2 + 8 = 0

0 = 0

x2 – 6x + 8 = 0

x2 – 6(4) + 8 = 0

0 = 0


Substitute for x.

Simplify.Solution checks.


2. x2 + x = – 1


Write original equation.x2 + x = – 1

Subtract – 1 from each side.x2 + x + 1 = 0

STEP 2

Graph the function y = x2 + x + 1 . The graph has no x-intercepts. The equation x2 + x = – 1 has no solution.

SOLUTION


3. – x2 + 6x = 9


Write original equation.– x2 + 6x = 9

Subtract 9 from each side.– x2 + 6x – 9 = 0

STEP 2Graph the function y = – x2 + 6x – 9 The x-intercepts is 3.

SOLUTION

ANSWER

The solution of the equation – x2 + 6x – 9 is 3.

EXAMPLE 6 Solve a multi-step problem

An athlete throws a shot put with an initial vertical velocity of 40 feet per second as shown.

a. Write an equation that models the height h (in feet) of the shot put as a function of the time t (in seconds) after it is thrown.

b. Use the equation to find the time that the shot put is in the air.

Sports

Solve a multi-step problemEXAMPLE 6

a. Use the initial vertical velocity and the release height to write a vertical motion model.

h = – 16t2 + vt + s Vertical motion model

h = – 16t2 + 40t + 6.5 Substitute 40 for v and 6.5 for s.

b. The shot put lands when h = 0. To find the time t when h = 0, solve 0 = – 16t2 + 40t + 6.5 for t.

To solve the equation, graph the related function h = – 16t2 + 40t + 6.5 on a graphing calculator. Use the trace feature to find the t-intercepts.


ANSWER

There is only one positive t-intercept. The shot put is in the air for about 2.6 seconds.


WHAT IF? In Example 6, suppose the initial vertical velocity is 30 feet per second. Find the time that the shot put is in the air.

6.


a. Use the initial vertical velocity and the release height to write a vertical motion model.


h = – 16t2 + 30t + 6.5 Substitute 40 for v and 6.5 for s.

b. The shot put lands when h = 0. To find the time t when h = 0, solve 0 = – 16t2 + 30t + 6.5 for t.

ANSWER There is only one positive t-intercept. The shot put is in the air for about 2.1 seconds.

Warm-Up – 10.4

Solve the equation

ANSWER X = –7

ANSWER X ≈14.1 and x≈-14.1


1. x= – 49√

2. x2 = 200

ANSWER 0.75 sec

Solve the following equation.

A ball is dropped from a height 9 feet above the ground. How long does it take the ball to hit the ground?

4.


3. 4x2 = 36

ANSWER x = ±3

Vocabulary – 10.4• Square Root

• What number multiplied by itself gives the original

• If a2 = b, then b is the square root of a.

• Perfect Square

• Integers that have integer square roots.

Notes – 10.4– Use Square Roots to solve Quad. Eqns.

• FINDING ROOTS OF QUAD. EQNS1. Factoring (GCM and binomials)2. Tables3. Graphing4. “What’s the goal of solvign every alg. Eqn.

you will ever see???1. What process did we use to do this??

•If the quadratic equations are “simple” (i.e. no “bx” term), the easiest solution may be SSADMEP.

Examples 10.4

EXAMPLE 1 Solve quadratic equations

Solve the equation.a. 2x2 = 8

SOLUTION

a. 2x2 = 8Write original equation.

x2 = 4 Divide each side by 2.

x = ± 4 = ± 2 Take square roots of each side. Simplify.

ANSWER The solutions are – 2 and 2.

Solve quadratic equationsEXAMPLE 1

b. m2 – 18 = – 18 Write original equation.

m2 = 0 Add 18 to each side..

The square root of 0 is 0.m = 0

ANSWER

The solution is 0.

Solve quadratic equationsEXAMPLE 1

c. b2 + 12 = 5Write original equation.

b2 = – 7 Subtract 12 from each side.

ANSWER

Negative real numbers do not have real square roots. So, there is no solution.

EXAMPLE 2 Take square roots of a fraction

Solve 4z2 = 9.

SOLUTION

4z2 = 9. Write original equation.

z2 = 94 Divide each side by 4.

Take square roots of each side.z = ± 94

z = ± 32

Simplify.

Take square roots of a fractionEXAMPLE 2

ANSWER

The solutions are – and 32

32

Approximate solutions of a quadratic equation

EXAMPLE 3

Solve 3x2 – 11 = 7. Round the solutions to the nearesthundredth.

SOLUTION

3x2 – 11 = 7 Write original equation.

3x2 = 18 Add 11 to each side.

x2 = 6 Divide each side by 3.

x = ± 6 Take square roots of each side.

Approximate solutions of a quadratic equation

EXAMPLE 3

x ± 2.45 Use a calculator. Round to the nearesthundredth.

ANSWER

The solutions are about – 2.45 and about 2.45.


Solve the equation.

1. c2 – 25 = 0

SOLUTION

c2 – 25 = 0 Write original equation.

c = ± 25 = ± 5 Take square roots of each side. Simplify.

GUIDED PRACTICE for Examples 1,2 and 3

ANSWER

The solutions are – 5 and 5.


Solve the equation.2. 5w2 + 12 = – 8

SOLUTION

5w2 + 12 = – 8 Write original equation.

w = –4 Take square roots of each side. Simplify.

GUIDED PRACTICE

ANSWER

Negative real numbers do not have a real square root.So there is no solution.

5w2 = – 8 –12 Subtract 12 from each side.

for Examples 1,2 and 3


Solve the equation.3. 2x2 + 11 = 11

SOLUTION

2x2 + 11 = 11 Write original equation.

x = 0 The root of 0 is 0.

GUIDED PRACTICE

ANSWER

The solution is 0 .

2x2 = 0 Subtract 11 from each side.



Solve the equation.4. 25x2 = 16

SOLUTION

25x2 = 16 Write original equation.

Take square roots of each side.

GUIDED PRACTICE

Divided each to by 25.x =1625

x = ± 1625

x = ± 4 5

Simplify.

ANSWER

The solution is – and . 4 5

4 5



Solve the equation.5. 9m2 = 100

SOLUTION

9m2 = 100 Write original equation.


GUIDED PRACTICE

Divided each to by 9.m =100 9

m = ± 100 9

m = ± 10 3

Simplify.

ANSWER

The solution is – and . 10 3

10 3



Solve the equation.6. 49b2 + 64 = 0

SOLUTION

49b2 + 64 = 0 Write original equation.


GUIDED PRACTICE

Divided each to by 9.

49b2 = – 64

b2 = –64 49

Subtract 64 from each side.

ANSWERNegative real numbers do have real square root. So there is no solution.

b = – 6449



Solve the equation. Round the solution to the nearest hundredth.

7. x2 + 4 = 14

SOLUTION

x2 + 4 = 14 Write original equation.


GUIDED PRACTICE

Use a calculation. Round to the nearest hundredth.

x2 = 10 Subtract 4 from each side.

ANSWER The solutions are about – 3.16 and 3.16.

x = 10+–

x = +– 3.16




8. 3k2 – 1 = 0

SOLUTION

3k2 – 1 = 0 Write original equation.


GUIDED PRACTICE

Use a calculation. Round to the nearest hundredth.

3k2 = 1 Add 1 to each side.

k = +– 0.58

k2 = 1 3

k = +– 13



EXAMPLE 4 Solve a quadratic equation

Solve 6(x – 4)2 = 42. Round the solutions to the nearesthundredth.

6(x – 4)2 = 42 Write original equation.

(x – 4)2 = 7 Divide each side by 6.

x – 4 = ± 7 Take square roots of each side.

7 x = 4 ± Add 4 to each side.

ANSWER

The solutions = 4 + 6.65 and 4 – 1.35.7 7

Solve a quadratic equation

EXAMPLE 4

CHECKTo check the solutions, first write the equation so that 0 is on one side as follows: 6(x – 4)2 – 42 = 0. Then graph the related function y = 6(x – 4)2 – 42. The x-intercepts appear to be about 6.6 and about 1.3. So, each solution checks.



10. 2(x – 2)2 = 18

SOLUTION

2(x – 2)2 = 18 Write original equation.



(x – 2)2 = 9

Add 2 to each side.


x – 2 = +– 9

ANSWER

The solutions are about – 1 and 5.

x = 2 + 3



11. 4(q – 3)2 = 28

SOLUTION

4(q – 3)2 = 28 Write original equation.



(q – 3)2 = 7

Add 3 to each side.

Divided each side by 4.

q – 3 = +– 7

q = 3 + 7–

ANSWER

The solutions are and = 5.65 and . = 0.35 3 + 7 3 – 7



12. 3(t – 5)2 = 24

SOLUTION

3(t – 5)2 = 24 Write original equation.



(t – 5)2 = 8

Add 5 to each side.


t – 5 = +– 8

t = 5 + 8–

ANSWER

The solutions are and – 2.17 and. = –7.83. 5 + 8– 5 – 8–

Solve a multi-step problem

EXAMPLE 5

During an ice hockey game, a remote-controlled blimp flies above the crowd and drops a numbered table-tennis ball. The number on the ball corresponds to a prize. Use the information in the diagram to find the amount of time that the ball is in the air.

Sports Event


EXAMPLE 5

STEP 1Use the vertical motion model to write an equation for the height h (in feet) of the ball as a function of time t (in seconds).

SOLUTION


EXAMPLE 5


h = – 16t2 + 0t + 45 Substitute for v and s.

STEP 2Find the amount of time the ball is in the air by substituting 17 for h and solving for t.


Write model.

Substitute 17 for h.


Divide each side by 16.

Take positive square root.

Use a calculator.

h = – 16t2 + 45

17 = – 16t2 + 45

– 28 = – 16t2

28 16

= t2

28 16

= t

1.32 t


ANSWER

The ball is in the air for about 1.32 seconds

EXAMPLE 1 Solve quadratic equationsGUIDED PRACTICE for Examples 4 and 5

WHAT IF? In Example 5, suppose the table-tennis ball is released 58 feet above the ground and is caught 12 feet above the ground. Find the amount of time that the ball is in the air. Round your answer to the nearesthundredth of a second.

13.


STEP 1

Use the vertical motion model to write an equation for the height h (in feet) of the ball as a function of time t (in seconds).

SOLUTION


h = – 16t2 + 0t + 45 Substitute for v and s.

STEP 2Find the amount of time the ball is in the air by substituting 12 for h and solving for t.


Write model.

Substitute 12 for h.


Divide each side by – 16.

Take positive square root.

Use a calculator.

h = – 16t2 + 58

12 = – 16t2 + 58

– 46 = – 16t2

46 16

= t2

46 16

= t

1.70 t

ANSWER

The ball is in the are for about 1.70 second.

Warm-Up – 10.6


1. Solve |x – 6| = 4.

2. Solve |x + 5| – 8 = 2.

ANSWER 2, 10

ANSWER –15, 5

1. Solve x2 + 5x = -6.

2. Solve by graphing:x2 + 5x = -5Round to the nearest tenth.

ANSWER x = -2 or x = -3

ANSWER x = -3.6 or x = -1.4

Vocabulary – 10.6• Quadratic Formula

• Formula that will provide the roots (real and “non-real”) for ALL quadratic equations

Notes – 10.6–Solve Quad. Eqns. With the Quadratic Formula

• The Quadratic Formula below will allow you to find the roots of ALL quadratic equations.•Use this when:

1. It isn’t easily factorable.2. The numbers are so big or small that using a

calculator is difficult b/c of the window settings.

3. You don’t know how else to do it!

Examples 10.6


What are the solutions of 3x2 + 5x = 8?

– 1 and – A 83 B – 1 and 8

3 C 1 and – 83 D 1 and 8

3

SOLUTION

Write original equation.3x2 + 5x = 8

Write in standard form.3x2 + 5x – 8 = 0

Quadratic formula ±x = – b b2 – 4ac2a


x = – 5± 52 – 4(3)(– 8)2(3)

Substitute values in the

quadratic formula: a = 3, b = 5,

and c = – 8.

Simplify.= – 5± 1216

Simplify the square root.= – 5 ± 116

The solutions of the equation are – 5 + 116 = 1 and – 5 – 11

6

= – 83

CThe correct answer is C. A B DANSWER


2x2 – 2 = x

2x2 – x – 7 = 0

Solve 2x2 – 7 = x.


Write in standard form.

Quadratic formulax =

b2 – 4ac+ ––b2a

– (– 1) –+ ( –1)2 – 4(2)(– 7)

2(2)=

Substitute values in the quadratic formula: a = 2, b = – 1, and c = – 7.

Simplify.

4= + –1 57


Write the equation in standard form, 2x2 – x – 7 = 0. Then graph the related function y = 2x2 – x – 7. The x-intercepts are about – 1.6 and 2.1. So, each solution checks.

ANSWER

571 + 57 2.14 and 1 – 4

– 1.64.4

The solutions are

CHECK

GUIDED PRACTICE for Examples 1and 2

Solve the equation by Quadratic Formula.x2 – 8x +16 = 0.1

SOLUTION

x2 – 8x +16 = 0 Write original equation.


b2 – 4ac+ ––b2a

Substitute values in the quadratic formula: a = 1, b = – 8, and c = 16.

Simplify.2

=+ –8 0

– – 8 –+ ( –8)2 – 4 1 16x=

2 1

ANSWER The solution of the equation is 4.


Solve the equation by Quadratic Formula.3n2 – 5n = – 1.2

SOLUTION

3n2 – 5n = – 1 Write original equation.

Substitute values in the quadratic formula: a = 3, b =( – 5), and c = 1.

Simplify.6

=+ –5 13

–(–5) –+ ( –5)2 – 4 3 1x=

2 1

3n2 – 5n +1 = 0 Write in standard form.


b2 – 4ac+ ––b2a


ANSWER The solutions are6

= 5 + 3 = 1 43and

6= 5 – 3 = 0 23

Write the equation in standard form, 3n2 – 5n + 1 = 0. Then graph the related function y = 3n2 –5n + 1 . The x-intercepts are about 0.23 and 1.4.3s So, each solution checks.

CHECK


Solve the equation by Quadratic Formula.4z2 = +7z +22

SOLUTION

4z2 = +7z +2 Write original equation.

Substitute values in the quadratic formula: a = 4, b = 7, and c = 2.

Simplify.8

=+ –7 81

–(–7) –+ ( –7)2 – 4 4 (2)x=

2 4

4z2 –7z – 2 = 0 Write in standard form.


b2 – 4ac+ ––b2a


ANSWER

The Solution are8

= 7 + 81 = 2.0and

8= 7 – 81 = 0.25

EXAMPLE 3 Use the quadratic formula

y = 10x2 – 94x + 3900

4200 = 10x2 – 94x – 3900

0 = 10x2 – 94x – 300

Write function.

Substitute 4200 for y.


SOLUTION

For the period 1971 2001, the number y of films produced in the world can be modeled by the function y = 10x2 94x + 3900 where x is the number of years since 1971. In what year were 4200 films produced?

–

–

Film Production

EXAMPLE 3 Use the quadratic formula

Substitute values in

the quadratic formula: a = 10, b = –94, and c = – 300.

20,83694 + –20

=Simplify.

The solutions of the equation are

ANSWER There were 4200 films produced about 12 years after 1971, or in 1983.

94 + 20,836

2012 and 94 – 20,836

20–3.

x = (–94)2 – 4 (10)(–300)–(–94)

+ –2(10)


WHAT IF? In Example 3, find the year when 4750 films were produced.

4.

y = 10x2 – 94x + 3900

4750 = 10x2 – 94x + 3900

0 = 10x2 – 94x – 850

Write function.

Substitute 4750 for y.


SOLUTION


Substitute values in

the quadratic formula: a = 10, b = –94, and c = – 850.

42,83694 + –20

=Simplify.

The solutions of the equation are

ANSWER There were 4750 films produced about 15 years after 1971, or in 1986.

94 + 42,836

2015 and 94 – 42,836

20–6.

8836

x = (–94)2 – 4 (10)(–850)–(–94)

+ –2(10)

EXAMPLE 4 Choose a solution method

Tell what method you would use to solve the quadraticequation. Explain your choice(s).

a. 10x2 – 7 = 0

SOLUTION

a. The quadratic equation can be solved using square roots because the equation can be written in the form x2 = d.

Choose a solution methodEXAMPLE 4

b. The equation can be solved by factoring because the expression x2 + 4x can be factored easily. Also, the equation can be solved by completing the square because the equation is of the form ax2 + bx + c = 0 where a = 1 and b is an even number.


b. x2 + 4x = 0

SOLUTION

Choose a solution methodEXAMPLE 4

c. The quadratic equation cannot be factored easily, and completing the square will result in many fractions. So, the equation can be solved using the quadratic formula.


c. 5 x2 + 9x – 4 = 0

SOLUTION



SOLUTION

Factoring because the expression factors easily.

x2 +x – 6 = 0 5.



SOLUTION

Factoring because the expression factors easily. Using square roots is another option since the equation can be written in the form x2 = d.

x2– 9 = 0 6.



SOLUTION

Completing the square because the equation is of the form

ax2 + bx = c where a = 1 and b is an even number. Another method is the quadratic formula since the equation does not factor easily.

x2 +6x =5 7.

Warm-Up – 10.7-10.8

ANSWER 18.5 ft

A ball is kicked into the air from a height of 4.5 feet with an initial velocity of 30 feet per second. What is the height of the ball after 1 second?

3.




1. Graph y = 3x.

ANSWER Linear

ANSWER

Tell whether the ordered pairs (0, 0), (1, 2), (2, 4), and (3, 6) represent a linear function, a quadratic function, or an exponential function. Why??

2.


ANSWER 5, –7, –5

For y = x2 – 3x – 5, find corresponding y-values for the x-values –2, 1, and 3.

3.

Vocabulary – 10.7-10.8• Discriminant

• Value of the expression b2 – 4(a)( c) in the quadratic formula

Notes – 10.7 – Interpret the discriminant.

Notes –10.8– Linear, Quad. And Exponential func’s.

To determine which function is represented, 1.graph the points and look at the shape,2.Check differences/patterns or3.Use the Regression functions on “Calli.”

Examples 10.7-10.8

Use the discriminant EXAMPLE 1

Number of solutions

Equationax2 + bx + c = 0

Discriminantb2 – 4ac

a. 2x2 + 6x + 5 = 0

b. x2 – 7 = 0

c. 4x2 – 12x + 9 = 0

62 – 4(2)(5) = –4 No solution

02 – 4(1)(– 7) = 28 Two solutions

(–12)2 –4(4)(9) = 0 One solution

Find the number of solutions EXAMPLE 2

Tell whether the equation 3x2 – 7 = 2x has two solutions, one solution, or no solution.

SOLUTION

STEP 1 Write the equation in standard form.

3x2 – 7 = 2x

3x2 – 2x – 7 = 0

Write equation.

Subtract 2x from each side.

Find the number of solutionsEXAMPLE 2

STEP 2 Find the value of the discriminant.

b2 – 4ac = (–2)2 – 4(3)(–7)

= 88

Substitute 3 for a, – 2 for

b, and –7 for c.

Simplify.

The discriminant is positive, so the equation has two solutions.

ANSWER

Tell whether the equation has two solutions, one

solution, or no solution.1. x2 + 4x + 3 = 0

SOLUTION

STEP 1 Write the equation in standard form.

x2 + 4x + 3 = 0 Write equation.


STEP 2 Find the value of the discriminant.

b2 – 4ac = (– 4)2 – 4(1)(3)

= 4

Substitute 1 for a, 4 for b, and 3 for c.

Simplify.

The discriminant is positive, so the equation has two solutions.

ANSWER


Fountains


The Centennial Fountain in Chicago shoots a water arc that can be modeled by the graph of the equation y= 0.006x2 + 1.2x + 10 where x is the horizontal distance (in feet) from the river’s north shore and y is the height (in feet) above the river. Does the water arc reach a height of 50 feet? If so, about how far from the north shore is the water arc 50 feet above the water?

SOLUTION


STEP 1

Write a quadratic equation. You want to know whether the water arc reaches a height of 50 feet, so let y = 50. Then write the quadratic equation in standard form.

Write given equation.y 0.006x2 1.2x 10 = – + +

50 0.006x2 1.2x 10= – + + Substitute 50 for y.

Subtract 50 from each side.0 0.006x2 1.2x 40= – + –


a = – 0.006, b = 1.2, c = – 40

Simplify

b2 4ac = (1.2)2 4(– 0.006)( 40)– – – = 0.48

STEP 3

Interpret the discriminant. Because the discriminant is positive, the equation has two solutions. So, the water arc reaches a height of 50 feet at two points on the water arc.

STEP 2

Find the value of the discriminant of0 = 0.006x2 + 1.2x – 40.


–x =

b 2 4ac– ±b2a

Quadratic formula

= – ±

1.2 0.48–2( 0.006)

Substitute values in the quadratic formula. 0

Use a calculator.x 42 or x 158

STEP 4

Solve the equation to find the distance from the north shore where the water arc is 50 feet above the water.

0 = 0.006x2 + 1.2x – 40.


The water arc is 50 feet above the water about 42 feet from the north shore and about 158 feet from the north shore.

ANSWER

Identify functions using differences or ratios EXAMPLE 2

Use differences or ratios to tell whether the table of values represents a linear function, an exponential function, or a quadratic function.

ANSWER

The table of values represents a quadratic function.

x –2 –1 0 1 2

y –6 –6 –4 0 6

First differences: 0 2 4 6

Second differences: 2 2 2

Identify functions using differences or ratiosEXAMPLE 2

ANSWER

The table of values represents a linear function.

x – 2 – 1 0 1 2

y – 2 1 4 7 10

Differences: 3 3 3 3


1. Tell whether the ordered pairs represent a linear function, an exponential function, or a quadratic function: (0, – 1.5), (1, –0.5),(2,2.5),(3,7.5).

ANSWER

quadratic function


2. Tell whether the table of values represents a linear function, an exponential function, or a quadratic function and find the function!

ANSWER exponential function

Y = 2 * 5x

0

y 2

x – 2 – 1 1

0.08 0.4 10

*5 *5 *5

Write an equation for a functionEXAMPLE 3

Tell whether the table of values represents a linear function, an exponential function, or a quadratic function. Then write an equation for the function.

x – 2 – 1 0 1 2

y 2 0.5 0 0.5 2

SOLUTION


STEP 1 Determine which type of function the table of values represents.

x – 2 – 1 0 1 2

y 2 0.5 0 0.5 2

First differences: – 1.5 – 0.5 0.5 1.5

Second differences: 1 1 1


STEP 2 Write an equation for the quadratic function. The equation has the form y = ax2. Find the value of a by using the coordinates of a point that lies on the graph, such as (1, 0.5).

y = ax2 Write equation for quadratic function.

0.5 = a(1)2 Substitute 1 for x and 0.5 for y.

0.5 = a Solve for a.

The table of values represents a quadratic function because the second differences are equal.


ANSWER

The equation is y = 0.5x2.

EXAMPLE 3GUIDED PRACTICE


STEP 1 Determine which type of function the table of values represents.

x – 3 – 2 – 1 0 1

y –7 –5 – 3 – 1 1

3.

for Example 3

SOLUTION

EXAMPLE 3GUIDED PRACTICE

x – 3 – 2 – 1 0 1

y –7 –5 – 3 – 1 1

– 2 – 2 – 2 – 2First difference

The table of value represent a linear function.

for Example 3

EXAMPLE 3GUIDED PRACTICE for Example 3

STEP 2 Write an equation for the quadratic function. The equation has the form y = ax2. Find the value of a by using the coordinates of a point that lies on the graph, such as (1, 1).

y = ax2 Write equation for quadratic function.

1= a(1)2 Substitute 1 for x and 0.5 for y.

1 = a Solve for a.

The equation is y = 2x - 1


x – 2 – 1 0 1 2

y 8 2 0 2 4

4.



x – 2 – 1 0 1 2

y 8 2 0 2 8

SOLUTION

Determine which type of function the table of values represents.

The table of values represents a quadratic function. The equation is y = 2x2

Review – Ch. 10 – PUT HW QUIZZES HERE

Daily Homework Quiz For use after Lesson 10.1

1. Graph y = –0.5x2 + 2

2. How would the graph of the function y = – 2x2 + 3 be affected if the function were changed to y = – 2x2 – 3?

ANSWER

It would be shifted down 6 units.ANSWER


3. A pinecone falls about 50 feet from the branch of a pine tree.Its height (in feet) can be modeled by the function h(t) = -16t2 + 50, where t is the time in seconds. How long does it take to land on the ground?

ANSWER about 1.8 sec


1. Find the axis of symmetry and the vertex of the graph of the function y = – 4x2 + 8x – 9

2. Graph y = –2x2 + 4x + 1

ANSWER x = 1; (1, – 5)

ANSWER


3. An arch to the entrance of the library can be modeled by y = – 0.13x2 + 2.5x, where x and y are measured in feet. To the nearest foot, what is the height of the highest point of the arch?

ANSWER 12 ft


1. Solve x2 +6x + 8 by graphing.

– 4 – 2ANSWER


2. x2 + 6x = – 10

ANSWER none

3. x2 + 6x = – 9

oneANSWER

Find the number of solutions for each equation.


4. Find the zeros of f(x) = – x2 + 2x + 3.

– 1, 3ANSWER

5. Approximate the zeros of f(x) = x2 + x – 3 to the nearest tenth.

ANSWER – 2.3, 1.3


6. Maria throws a shot put with an initial velocity of 25 feet per second. She releases it at a height of 5 feet. Find the time the shot put is in the air.

about 1.7 secANSWER

ANSWER –2, 2


Solve the equation. Round solution to the nearest hundredth, if necessary.

1. 4b2 – 13 = 3

2. 9x = 25

ANSWER –3.16, 3.16

ANSWER – 53

53

,

3. 3n2 –18 = 12


At a football game you are sitting 32 feet above the ground. If your hat comes off falls to the ground, how long will it be in the air?

4.

ANSWER About 1.41 sec

ANSWER – 1.29, 9.29

ANSWER –14, 2


1. x2 + 12x = 28

Solve the equation by completing the square. Round to the nearest hundredth, if necessary.

2. m2 – 8m = 12


What is the width of the border that surrounds this poster?

3.

ANSWER 1 in


1. 6x2 – 6 = – 5x

Use the quadratic formula to solve the equation.Round your solutions to the nearest hundredth, if necessary.

2. 2x2 + 3x – 8 = 0

–2.87, 1.39ANSWER

,ANSWER32

– 23


3. 3m2 – m + 9 = 12

ANSWER – 0.85, 1.18

You have seen that the function y = 10x2 + 94x + 3900 models the number y of films produced in the world, where x is the number of years since 1971. In what year were 10,000 films produced?

4.

ANSWER about 29.8 years after 1971, or during the year 2000


1. 4b2 + 2b – 5 = 0

Tell whether the equation has two solutions, one solution, or no solutions.

2. 2g2 + 8g = – 11


3. y = x2 + 14x + 49

y = x2 + 14x + 504.

Find the number of x-intercepts of the graphs of the equations


5. The graph of y = – 0.2x2 + 3.5x models the height of one of the arches at the entrance to a parking structure. Can a truck that is 20 feet high fit under the arch?

Daily Homework Quiz For Section 10.8

6. Tell whether the table of values represents a linear function, an exponential function, or a quadratic function and find the function!

1

y 1

x – 1 0 2

16 4 1/4

2) ANSWER no solutions

1) ANSWER two solutions

Answers

3) ANSWER one

4) ANSWER none

The value of the discriminate of 0 = – 0.2x2 + 3.5x – 20 is negative, so there is no solution.The truck will not fit under the arch.

5) ANSWER

6) ANSWER exponential function

Y = 4(1/4)x

Warm-Up – X.X

Vocabulary – X.X• Holder

• Holder 2

• Holder 3

• Holder 4

Notes – X.X – LESSON TITLE.• Holder•Holder•Holder•Holder•Holder

Examples X.X

Documents

Unit 6 – Chapter 10