Algebra 1 Unit 2 Lesson 6 to 10

Lesson 6 Quadratic Functions

parabola the u-shaped graph of a quadratic function

quadratic equation

an equation that can be written as ax² + bx + c = 0, where the leading coefficient, a, is greater than or equal to one

quadratic function

a function that can be written as f(x) = ax² + bx + c, where "a" is not zero

Please take note here:

A quadratic function is of the form f(x) = ax 2 + bx + c , where "a" is not zero. The graph of a quadratic function is a parabola. A quadratic equation can be written in the form ax 2 + bx + c = 0, where the

leading coefficient, a, is greater than or equal to one A quadratic equation is in standard form when "a" is positive and a, b, and c are

integers whose greatest common factor is 1. The solutions of 0 = ax 2 + bx + c are the x-intercepts of the graph of y = ax 2 + bx

+ c.

Quadratic Functions:

A quadratic function is a function that can be written in the standard form f(x) = ax 2 + bx + c, where "a" is not zero. Recall that in function notation, f(x) represents y, so we may write y = ax 2 + bx + c.

Quadratic EquationsMuch of the work we will do with quadratic functions deals with the related quadratic equation.

A quadratic equation is an equation that can be written in the form y= ax  2 + bx + c = 0, where "a" is not zero.

A quadratic equation of the form y = ax 2 + bx + c = 0, where "a" is not zero, is said to be in standard, or general, form when:

1. "a" is positive;2. a, b, and c are integers whose greatest common factor is 1.

Example 1:2x 2 + 3x - 4 = 0 is a quadratic equation in standard form.a = 2b = 3c = -4 The terms cannot be reduced as the largest factor common to 2, 3, and -4 is 1.

Example 2:x 2 + 7 = 0 is a quadratic equation in standard form.a = 1b = 0 (There is no "x" term.)c = 7 The largest common factor is 1.

Example 3:-5x 2 + x = 0 is a quadratic equation. However, a is negative; therefore, this is not in standard form. The standard form is found by multiplying both sides of the equation by -1.-1[-5x 2 + x] = -1[0]5x 2 - x = 0a = 5b = -1c = 0The largest common factor is 1, so 5x 2 - x = 0 is in standard form.

Example 4:2x 2 - 4x + 6 = 0 is a quadratic equation. 2 is a common factor in a, b, and c. Therefore, the standard form is found by dividing all the terms by 2.

 a = 1b = -2c = 3

Example 5:0.7x 2 = 1 is a quadratic equation. a is not an integer, and the equation does not equal zero, so this is not in standard form.Multiplying by 10 will move the decimal one place.10[0.7x 2] = 10[1]7x 2 = 10Set the equation equal to zero:7x 2 - 10 = 0a = 7b = 0c = -10

Example 6:

(x + 2)(x - 7) = 5

3[ (x + 2)(x - 7)] = 3[5]1(x + 2)(x - 7) = 15x 2 - 5x - 14 = 15x 2 - 5x - 29 = 0a = 1b = -5c = -29Remember, to be a quadratic equation, the integers a, b, and c must have a greatest common factor of 1.

Example 7:x 3 - 2x 2 + 1 = 0 is not a quadratic equation because it contains a third-degree term.

Example 8:3x - 11 = 0 is not a quadratic equation because it does not contain a second-degree term. In fact, you may recall that this is a linear equation.

Solve Quadratic Equations Graphically

The solution of the quadratic equation is the point or points on the graph where y is equal to zero, or the x-intercepts. If the parabola never crosses the x-axis, there is no solution to the equation. 

Example:The parabola shown is the graph of y = x  2 - 4x + 5.

 

Find the solutions of the quadratic equation 0 = x  2 - 4x + 5.Solution:

The solutions are the x-intercepts.The parabola does not cross the x-axis. There are no solutions to the quadratic equation.The solution set of 0 = x 2 - 4x + 5 is Ø.

Lesson 7 Transformations

reflectiona transformation resulting in a mirror image over a line

standard form the form y = ax2 + bx + c, where a is an integer greater than zero

translation a shift or slide of a graph in the coordinate plane

vertexthe turning point of a parabola; the minimum or maximum point on the parabola

Please take note:

For any natural number c:

A parabola has a line of symmetry. The vertex lies on the line of symmetry. Equations of the form y = x 2 + k are vertical translations of k units from y = x 2.

If k is positive, the graph moves up k units; if k is negative, the graph moves down k unit. The line of symmetry remains the same.For example:y = x 2 + 2 translates 2 units up from y = x 2 and y = x 2 -2 translates 2 units down from y = x 2

Equations of the form y = (x - h) 2 are horizontal translations of y = x 2.If h is positive, the graph moves right h units; if h is negative, the graph moves to the left h units. The line of symmetry changes.For example:y = (x + 2) 2 translates 2 units to the left from y 2 and y = (x - 2) 2 translates 2 units to the right from y = x 2

The vertex form of a quadratic equation is y = (x - h) 2 + k, where (h, k) is the vertex and x = h is the line of symmetry.

Make note of some things about the graph:

1. The graph is a parabola because the function is quadratic. (If we did not know this, then

we would need many more points to determine the shape.)

2. The domain (set of x-values) is all real numbers because we can square any number.

Therefore, we know that the points are connected with a solid line.

3. The range (set of y-values) must be greater than zero because any real number squared

will result in a positive value. Therefore, the graph continues upward. The range is y ≥ 0.4. The graph is symmetric with respect to the y-axis because, for all real numbers, x 2 is

equal to (-x) 2.All parabolas have a turning point called the vertex and a line of symmetry. In the y = x 2 graph, the vertex is the point (0, 0), and the line of symmetry is the y-axis (x = 0).The process of graphing a function using a table of values works, but there are some disadvantages. Not only can it be time consuming to find several ordered pairs, but it can be difficult to know where the vertex is located. By randomly choosing x values, you may end up with points that are all on one side of the line of symmetry.Understanding the relationship between reflecting and shifting the graph of y = x 2 and its equation can make the task of graphing parabolas easier.Reflecting over the x-axis. Examine the table shown below for the graph of y = -x 2.Reminder:

-x 2 says to take the negative of x 2.Square the x value first and then negate it.

x ƒ(x) = -x 2 ƒ(x)

- 3 ƒ(x) = -(-3) 2 -9

- 2 ƒ(x) = -(-2) 2 -4

- 1 ƒ(x) = -(-1) 2 -1

0 ƒ(x) = -(0) 2 0

1 ƒ(x) = -(1) 2 -1

2 ƒ(x) = -(2) 2 -4

3 ƒ(x) = -(3) 2 -9

When the coefficient of x 2 (the "A") is negative, the parabola opens down.Vertical translations. A vertical translation of a graph in the coordinate plane is a shift of the graph up or down. Graphs of quadratic equations of the form y = x 2 + k are vertical translations of the graph of y = x 2. Notice in the examples shown that the value of k changes the y-coordinate.When the equation is of the form y = x 2 + k, the graph is a vertical shift k units of y = x 2

Note: Depending on how the graph is moved this will tell how many solutions the function will have. Recall that quadratic functions have one of the following solutions: one real solution, two real solutions or no real solutions.

Example #1:Graph y = x 2 + 2. Tell how many solutions the graph will have.This is a vertical shift of y = x 2 up 2 units.

Note: When moving the graph 2 units up, we no longer have x-intercepts. Therefore, this function will have NO REAL SOLUTIONS.Note that, for vertical translations, the line of symmetry is still x = 0. The vertex is (0, k).When the equation is of the form y = x 2 + k, the graph is a vertical shift k units of y = x 2.Horizontal translations. A horizontal translation of a graph in the coordinate plane is a shift of the graph left or right. Graphs of quadratic equations of the form y = (x + h) 2 are horizontal translations of the graph of y = x 2. Notice in the examples shown that the value of h changes the x-coordinate. Note that, for horizontal translations, the line of symmetry is x = h. The vertex is (h, 0).When the equation is of the form y   = ( x   -   h )   2 , the graph is a horizontal shift   h   units of   y   =   x   2 .

Example #3:Graph y = (x + 2) 2. Tell how many solutions the graph will have.This is a horizontal shift of y = x 2 to the left 2 units.

Since the graph was shifted left, the graph will have 1 x-intercepts. Therefore, this function will have ONE REAL SOLUTION.

Lesson 8 Transformations: Exponential Equations

As you have previously learned, two distinct points can be used to identify linear equations. You also discovered that you need more points to graph a quadratic function because the graph is a curve. Similarly, you will need a table of values to graph the curve for an exponential function. Graphical inspection is a quick solution method, and it can also be used to graph transformations and identify equations associated with transformations of exponential functions.

asymptote a line that a curve approaches but never reaches when values of x approach a certain number

reflection a transformation resulting in a mirror image over a line

translation a shift or slide of a graph in the coordinate plane

Please take note:

The domain of an exponential function is the set of all real numbers. The asymptote of an exponential function is found by inspecting the function

values when the exponent takes on large positive or negative values. The graph of an exponential function may be translated vertically or horizontally,

reflected across the x-axis or y-axis, and stretched vertically or horizontally.

With a quadratic function, it was helpful to find the vertex and decide whether the

function opened up or down. With an exponential function, identifying the asymptote is one key to making a graph. It

also helps to find the x-value that yields an exponent of 0, which gives ƒ(x) = a0 = 1. Then, choose x-values near this value when making a table.

Take Note of the following transformation (Reflection, Translation, Dilation):Reflection

Translation

DilationDilating an exponential function vertically. 

You have looked at a special case of ƒ(x) = h•ax, when h = -1 the curve is reflected over the x-axis. When h > 1, the curve is dilated, or stretched, vertically with no reflection across the x-axis. When h < -1, the curve is stretched vertically and also is reflected across the x-axis.

Example1 :Graph g(x) = 2•2 x .

Solution:Multiplying 2 x by 2 represents a vertical stretch by a factor of two.

Dilating an exponential function horizontally. You have seen that ƒ(x) = ahx represents a reflection over the y-axis when h = -1. When h > 1, the curve is stretched horizontally with no reflection across the y-axis. When h < -1, the curve is stretched vertically and also is reflected across the y-axis.

Example 2:Graph g(x) = 2-2x.

Solution:Multiplying x by -2 stretches the curve vertically and also reflects it over the y-axis.

Composite functions. Some functions include more than one transformation. Here’s an example.

Example 3:What transformations of ƒ(x) = 2x are represented by g(x) = -3⋅22x?

Solution:Multiplying the exponent by 2 indicates a horizontal stretch by a factor of 2.

Multiplying the exponential expression by -3 indicates a vertical stretch by a factor of 3 and a reflection over the x-axis.Therefore, this function represents a vertical dilation, a horizontal dilation, and one reflection.

Lesson 9 Line of Symmetry

Take Note:

The formula for the line of symmetry is  . To find the vertex, substitute the value of x for the line of symmetry into the

equation to find y. Steps for graphing:

1. Determine if the parabola opens up or down.2. Find the line of symmetry.3. Find the vertex.4. Use a table of values to plot a couple of points on each side of the line of

symmetry.5. Connect the points with a smooth curve.

Take Note:

EXAMPLE:

Consider graphing the equation y = x2 + 6x + 9. You may recognize the trinomial x2 + 6x + 9 as a perfect square. It factors into (x + 3)(x + 3) or (x + 3)2.So, we have that y = x2 + 6x + 9 can be written as y = (x + 3)2. From this standard form, we can tell that the line of symmetry is x = -3 and the vertex is (-3, 0).This method works fine if the trinomial is a perfect square. But consider graphing y = x2 + 8x + 5. This trinomial is not a perfect square. In fact, it is not even factorable. In these cases, we need another method for graphing.

Fortunately, there is an equation that allows us to find the line of symmetry, and therefore the vertex, of a parabola. This equation gives us a starting point for graphing. With it, we know where to start choosing x values to get points for the graph.

The equation for the line of symmetry of a parabola whose equation is y = ax2 + bx + c is:

Practice:1. Find the line of symmetry and vertex for the graph of y = x2 + 6x - 1.2. Find the line of symmetry and vertex for the graph of y = x2 - x - 7.

Practice:

Graph:

Lesson 10 Quadratics Inequalities

Take Note:

A point is a solution of a quadratic inequality if it makes the statement true. To graph a quadratic inequality:

1. Write the quadratic inequality in standard form.2. Graph the related quadratic equation to get the boundary.3. Make the boundary solid if the inequality includes "or equal"; make the

boundary dashed if it does not.4. Shade below the boundary if y is less than; shade above if y is greater

than.

Practice: