Choosing a Method of SolutionAlthough you can use the quadratic formula to solve any equation, it is often much easier to factor or complete the square.

The list below suggests when to use which method.

Situation Method to use

2, , and are integers and 4 is a perfect square.a b c b ac factoring

2

The equation has the form

even number constant 0x x completing the square

2

In the remaining situations the quadratic formula is usually easiest to use,

especially when solving an equation like 0, where the coefficients

are letters. Also the quadratic formula is the mo

px qx r st efficient method to use when

approximating real roots of a quadratic equation using a calculator.

If an equation contains variables on both sides or variables in the denominator, then you must carefully organize your method for solving in order not to lose a root or gain a root.

It is possible to lose a root by dividing both sides of an equation by a common factor.

24 1 3 1x x x divide both sides by 1 . x

4 3 1x x 4 3 3x x

3x 2

Actually the roots of 4 1 3 1 are 1 and 3.x x x

When we divided both sides of the equation by 1, we lost a root.x

Two ways to avoid losing a root are shown below. Both methods are correct.

Method 1 Method 2

If there is a factor common to both sides of the equation, remember to include as roots all values that make this factor zero.

24 1 3 1x x x

If 1 0, then

1 and both

sides of the equation

equal to zero.

So, 1 is a root.

x

x

If 1 0, we

can divide both

sides by 1, getting

4 3 1 .

So, 3 is a root.

x

x

x x

Bring all terms to one side of the equation and then solve.

24 1 3 1x x x

24 1 3 1 0x x x 24 1 3 2 1 0x x x x

2 24 4 3 6 3 0x x x x 2 2 3 0x x

1 3 0x x

1 or 3x x

Gaining a root

Squaring both sides or multiplying by an expression may give you an extraneous root which satisfies the transformed equation but not the original one.

2

2 2 8 4

2 2 4

x x xSolve

x x x

Multiply each side by the LCD

2 2 2 2 8 4x x x x x 2 24 4 4 4 8 4x x x x x

22 8 8 4x x 22 4x x 22 4 0x x

2 2 0x x 2 0 2 0x x

0 2x x

Check:

0x

2

8 4 00 2 0 2

0 2 0 2 0 4

2 2 8

2 2 4

1 1 2

0so x

2x

2

8 4 22 2 2 2

2 2 2 2 2 4

0 4 0

4 0 0

Undefined, so x = 2 is not a solution.

Ex. 1. The senior class has paid $200 to rent a roller skating rink for a fund raising party. Tickets for the party are $5 each.

a. Express the net income as a function of the number of tickets sold.

b. Graph the function. Identify the point at which the class begins to make a profit.

a. Let the net income when tickets are sold.I n n

Net income is the amount of money remaining after operating expenses are deducted from the amount of money earned.

$5 times the number of tickets sold minus $200 rental feeI n

5 200I n n

Since n must be a nonnegative integer, the graph consists of discrete (isolated) points on a line.

However, the line that contains the discrete points is often given as a sketch of the function.

From these graphs you can see that n = 40 is a zero of the income function.

Therefore, the class must sell more than 40 tickets to make a profit.

The domain of a function is the set of values for which the function is defined.

In the function 5 200 in example 1, represents the number of tickets sold,

so the domain is the set of nonnegative integers.

I n n n

Substituting 3 or 4.5 does not make sense in this case.n n

You can think of the domain of a function as the set of input values.

The set of output values is called the range of the function.

When you substitute the domain values 0,1,2,3,...,

you get the range values 200, 195, 190, 185,....

n

I n

A mathematical model is one or more functions, graphs, tables, equations, or inequalities that describe a real-world situation.

Example 2 illustrates another model

The function 5 200 is said to model income.I n n

Ex. 2. Suppose that it costs 50 cents for the first minute of a long distance telephone call and 20 cents for each additional minute or fraction thereof. Give a graphical model of the cost of a call lasting t minutes.

It is important to realize that the cost of a call lasting 2 minutes and 12 seconds is the same as a 3 minute call.

Likewise, a three and a half minute call costs the same as a 4 minute call.

The function that models this cost is a step function, so named because its graph has steps.

Because the linear function 50 20

gives good approximations of the costs, it can

be used to model the cost of long distance calls.

f x x

Be aware that if you use the linear function, you will get overestimates of the cost.

Quadratic Functions and their Graphs

2

2

The graph of the quadratic function ,

where 0, is the set of points , that satisfy the

equation +bx+c.

f x ax bx c

a x y

y ax

This graph is a parabola.

If a graph has an axis of symmetry, then when you fold the graph along this axis, the two halves of the graph coincide.

The graph of a quadratic function has a vertical axis of symmetry, or axis.

The vertex of the parabola is the point where the axis of symmetry intersects the parabola.

If 0, the parabola opens upward, and the function has a minimum value.a

If 0, the parabola opens downward, and the function has a maximum value.a The bigger is, the narrower the parabola is.a

2 2Notice that the graph of 3 is narrower than the graph of .y x y x

2The -intercept of a parabola with equation is .y y ax bx c c

2The -intercepts are the real roots of 0.x ax bx c

2

Since the quadratic equation may have two, one, or no real roots,

depending on the value of the discriminant 4 , we have three

possibilities regarding the -intercepts of a parabola.

b ac

x

The axis of symmetry is a vertical line midway between the -interceptsx

The equation of the axis is .2

bx

a

To sketch the graph of 2y ax bx c

•Decide if it opens up or down and the number of x-intercepts

•Find the axis and vertex

•Find the x- and y-intercepts

Ex. 1. Sketch the parabola .

Label the intercepts, axis of symmetry, and vertex.

22 8 5y x x

Does it open up or down? 2 0 opens up

Find the axis2

bx

a

8

2 2x

2 2

To find the vertex, plug in 2 for x and determine the y-coordinate.

22 2 8 2 5 3y vertex is 2, 3

Find the x and y-intercepts

intercept = 5y c -intercepts are the zeros of the functionx

28 8 4 2 5

2 2x

8 64 40

4

8 24

4

8 2 6

4

62

2

2

If the equation of a parabola is written in the form

,

then the vertex is , .

y a x h k

h k

2The vertex of the parabola 2 3 7 is 3,7 .y x

2The vertex of the parabola 4 9 2 is 9,2 .y x

Ex. 2.

a. Find the vertex of the parabola by completing the square.

b. Find the x- and y-intercepts22 12 4y x x

a. Put in vertex form: where (h, k)is the vertex 2y a x h k

22 12 4y x x 22( 6 ) 4x x

22 x 3

9 18

22

The vertex is (3, 22)

b.

When x = 0, y = 4. So the y-intercept is (0, 4)

To find the x-intercept, let y = 0

22 3 22y x

20 2 3 22x

222 2 3x

211 3x 11 3x

3 11 x

Ex. 3. Where does the line y = 3x + 5 intersect the parabola 29y x

Set and solve for x 23 5 9x x 23 5 9x x

2 3 4 0x x 4 1 0x x

4 0 1 0x x 4 1x x

Substitute these into y = 3x + 5 to get y = -7 and y = 8

So the intersection points are (-4, -7) and (1, 8)

Graph to confirm your answer

Ex. 4.

Find an equation of the function whose graph is a parabola with x-intercepts (3, 0) and (6, 0) and y-intercept (0, -2).

If x = 3 and x = 6 are solutions of this equation, then (x – 3) and (x – 6) are factors of the equation so

3 6y a x x Use the y-intercept

2 0 3 0 6a 2 3 6a

2 18a 1

9a

13 6

9y x x 21

9 189x x 21

29x x

212

9f x x x

Graph to confirm