homework 7 graphing rational functions

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Integrated math 3

Course: integrated math 3   >   unit 13.

• Graphing rational functions according to asymptotes
• Graphs of rational functions: y-intercept
• Graphs of rational functions: horizontal asymptote
• Graphs of rational functions: vertical asymptotes
• Graphs of rational functions: zeros
• Graphs of rational functions
• Graphs of rational functions (old example)

Graphing rational functions 1

• Graphing rational functions 2
• Graphing rational functions 3
• Graphing rational functions 4

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9.6: Graphs of Rational Functions

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• Page ID 45132

• Darlene Diaz
• Santiago Canyon College via ASCCC Open Educational Resources Initiative

Previously, in the chapters where we discussed functions, we had a function from the library \(f(x) = \dfrac{1}{x}\). Recall, the graph of this function is

We plotted some points we obtained from the table and determined that the domain is all real numbers except for \(x = 0: \{x|x\neq 0\}\) or \((−∞, 0) ∪ (0, ∞)\). We called this function a rational function .

Definition: Rational Function

A rational function , \(R(x)\), is a ratio of two polynomials, \(P(x)\) and \(Q(x)\), of the form \[R(x)=\dfrac{P(x)}{Q(x)},\nonumber\] where \(Q(x)\neq 0\).

In this textbook, we only discuss when \(P(x) = 1\) and when \(Q(x)\) is of the form \(x^n\), where \(n\) is a positive integer and \[R(x)=\dfrac{1}{x^n}\nonumber\] For cases when \(P(x)\) is a polynomial other than the constant function \(1\) and \(Q(x)\) is a polynomial other than the power function \(x^n\) is left for future Algebra classes.

Let’s investigate these functions a little further. We know the domain is all real numbers except for \(x = 0\), but let’s look at the graph more closely. Notice, in the graph of \(f(x)\) above, the graph doesn’t intersect the \(y\) axis. Why? Well, let’s set \(y = f(x) = 0\) and solve:

\[f(x)=0=\dfrac{1}{x}\nonumber\]

When is this fraction zero? We know from previous sections that a fraction is zero when the numerator is zero. Will the numerator ever be zero, i.e.,

\[1\stackrel{?}{=}0\nonumber\]

No, never! This means there are no values of \(x\) such that \(y = 0\), and that \(y = 0\) is not in the range of the function.

This is not a coincidence. The fact that \(x\neq 0\) and \(y\neq 0\) for the function \(f(x) =\dfrac{1}{x}\) means that \(f(x)\) has vertical and horizontal asymptotes at \(x = 0\) and \(y = 0\), respectively.

Definition: Horizontal Asymptote & Vertical Asymptote

A function, \(R(x)\), has a horizontal asymptote at \(y = 0\) and a vertical asymptote at \(x = 0\) when \(R(x)\) is of the form \[\dfrac{1}{x^n}\nonumber\] We denote these asymptotes by drawing dashed lines for lines \(x = 0\) and \(y = 0\).

Example 9.6.1

Let’s regraph \(f(x) = \dfrac{1}{x}\) showing the horizontal and vertical asymptotes at \(y = 0\) and \(x = 0\), respectively.

So, we see the asymptotes, in red, are dashed lines on the \(x\) and \(y\)-axis and are the lines \(y = 0\) and \(x = 0\). The only case in which the horizontal and vertical asymptotes move left or right, and up or down, respectively, is if there are shifts to the parent function \(f(x)\).

Example 9.6.2

Graph \(R(x)=\dfrac{1}{x^2}\).

Let’s pick \(x\)-coordinates, and find corresponding \(y\)-values.

Plot the ordered-pairs from the table. To connect the points, be sure to connect them from smallest \(x\)-value to largest \(x\)-value, i.e., left to right. The domain of \(R(x)\) is all real numbers except for \(x = 0: \{x|x\neq 0\}\) or \((−∞, 0)∪(0, ∞)\). Since \(R(x)\) has horizontal and vertical asymptotes at \(y = 0\) and \(x = 0\), respectively, let’s draw the lines that represent these asymptotes.

Graphing Rational Functions Using Shifts

Let’s take a look when rational functions’ graphs contain horizontal and vertical shifts. It will be interesting to see the horizontal and vertical asymptotes change due to these shifts.

Horizontal and Verticals Shifts of Rational Functions

Given \(R(x)\) is a rational function, a horizontal shift and vertical shift of \(R(x)\) are described below:

Example 9.6.3

Graph \(R(x)=\dfrac{1}{x-2}\).

Let’s start by taking the parent function \(f(x) = \dfrac{1}{x}\). We see that \(R(x) = f(x − 2)\) because we replaced \(x\) with the factor \((x − 2)\). Looking at the table above, we see this is a horizontal shift with \(h = 2\), moving \(2\) units to the right, and the vertical asymptote changes to \(x = 2\).

We can see the gray graph, \(f(x)\), moved two units to the right, in addition to the vertical asymptote. Recall, from the table above, the horizontal asymptote stays \(y = 0\). Hence, the blue graph, \(R(x)\), is the final graph after applying the shifts.

Example 9.6.4

Graph \(K(x)=\dfrac{1}{x}+1\).

Let’s start by taking the parent function \(f(x) = \dfrac{1}{x}\). We see that \(K(x) = f(x) + 1\) because we added \(1\) to \(f(x)\). Looking at the table above, we see this is a vertical shift with \(k = 2\), moving \(1\) unit upward, and the horizontal asymptote changes to \(y = 1\).

We can see the gray graph, \(f(x)\), moved one unit upward, in addition to the horizontal asymptote. Recall, from the table above, the vertical asymptote stays \(x = 0\). Hence, the blue graph, \(K(x)\), is the final graph after applying the shifts.

Example 9.6.5

Graph \(Q(x)=\dfrac{1}{x+1}-2\).

Let’s start by taking the parent function \(f(x) = \dfrac{1}{x}\). We see that \(Q(x) = f(x + 1) − 2\) because we replaced \(x\) with the factor \((x + 1)\) and we subtracted \(2\) from \(f(x)\). Looking at the table above, we see \(Q(x)\) has a few shifts: a horizontal shift with \(h = −1\), moving \(1\) unit to the left, a vertical shift with \(k = −2\), moving \(2\) units downward, and the vertical and horizontal asymptotes change to \(x = −1\) and \(y = −2\), respectively.

We can see the gray graph, \(f(x)\), moved one unit to the left and \(2\) units downward in addition to the horizontal and vertical asymptotes. Notice, we had a vertical and horizontal shift. We moved \(f(x)\) one unit left, then \(2\) units down for all points. These shifts cause the asymptotes to move too. In fact, the vertical asymptote moved one unit to the left and the horizontal asymptote moved \(2\) units downward. Hence, the blue graph, \(Q(x)\), is the final graph after applying the shifts.

Graphs Rational Functions Homework

Graph each rational functions using the parent function \(f(x) = \dfrac{1}{x}\). Include the vertical and horizontal asymptotes.

Exercise 9.6.1

\(R(x)=\dfrac{1}{x-1}\)

Exercise 9.6.2

\(Q(x)=\dfrac{1}{x+3}\)

Exercise 9.6.3

\(S(x)=\dfrac{1}{x}+2\)

Exercise 9.6.4

\(T(x)=\dfrac{1}{x}-4\)

Exercise 9.6.5

\(U(x)=\dfrac{1}{x+2}-1\)

Exercise 9.6.6

\(U(x)=\dfrac{1}{x-3}-2\)

Exercise 9.6.7

\(P(x)=\dfrac{1}{x}+3\)

Exercise 9.6.8

\(N(x)=\dfrac{1}{x-1}-4\)

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9.7 Graph Quadratic Functions Using Transformations

Learning objectives.

• Graph quadratic functions of the form f ( x ) = x 2 + k f ( x ) = x 2 + k
• Graph quadratic functions of the form f ( x ) = ( x − h ) 2 f ( x ) = ( x − h ) 2
• Graph quadratic functions of the form f ( x ) = a x 2 f ( x ) = a x 2
• Graph quadratic functions using transformations
• Find a quadratic function from its graph

Be Prepared 9.7

Before you get started, take this readiness quiz.

• Graph the function f ( x ) = x 2 f ( x ) = x 2 by plotting points. If you missed this problem, review Example 3.54 .
• Factor completely: y 2 − 14 y + 49 . y 2 − 14 y + 49 . If you missed this problem, review Example 6.24 .
• Factor completely: 2 x 2 − 16 x + 32 . 2 x 2 − 16 x + 32 . If you missed this problem, review Example 6.26 .

Graph Quadratic Functions of the form f ( x ) = x 2 + k f ( x ) = x 2 + k

In the last section, we learned how to graph quadratic functions using their properties. Another method involves starting with the basic graph of f ( x ) = x 2 f ( x ) = x 2 and ‘moving’ it according to information given in the function equation. We call this graphing quadratic functions using transformations.

In the first example, we will graph the quadratic function f ( x ) = x 2 f ( x ) = x 2 by plotting points. Then we will see what effect adding a constant, k , to the equation will have on the graph of the new function f ( x ) = x 2 + k . f ( x ) = x 2 + k .

Example 9.53

Graph f ( x ) = x 2 , g ( x ) = x 2 + 2 , f ( x ) = x 2 , g ( x ) = x 2 + 2 , and h ( x ) = x 2 − 2 h ( x ) = x 2 − 2 on the same rectangular coordinate system. Describe what effect adding a constant to the function has on the basic parabola.

Plotting points will help us see the effect of the constants on the basic f ( x ) = x 2 f ( x ) = x 2 graph. We fill in the chart for all three functions.

The g ( x ) values are two more than the f ( x ) values. Also, the h ( x ) values are two less than the f ( x ) values. Now we will graph all three functions on the same rectangular coordinate system.

The graph of g ( x ) = x 2 + 2 g ( x ) = x 2 + 2 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted up 2 units.

The graph of h ( x ) = x 2 − 2 h ( x ) = x 2 − 2 is the same as the graph of f ( x ) = x 2 f ( x ) = x 2 but shifted down 2 units.

Try It 9.105

ⓐ Graph f ( x ) = x 2 , g ( x ) = x 2 + 1 , f ( x ) = x 2 , g ( x ) = x 2 + 1 , and h ( x ) = x 2 − 1 h ( x ) = x 2 − 1 on the same rectangular coordinate system. ⓑ Describe what effect adding a constant to the function has on the basic parabola.

Try It 9.106

ⓐ Graph f ( x ) = x 2 , g ( x ) = x 2 + 6 , f ( x ) = x 2 , g ( x ) = x 2 + 6 , and h ( x ) = x 2 − 6 h ( x ) = x 2 − 6 on the same rectangular coordinate system. ⓑ Describe what effect adding a constant to the function has on the basic parabola.

The last example shows us that to graph a quadratic function of the form f ( x ) = x 2 + k , f ( x ) = x 2 + k , we take the basic parabola graph of f ( x ) = x 2 f ( x ) = x 2 and vertically shift it up ( k > 0 ) ( k > 0 ) or shift it down ( k < 0 ) ( k < 0 ) .

This transformation is called a vertical shift.

Graph a Quadratic Function of the form f ( x ) = x 2 + k f ( x ) = x 2 + k Using a Vertical Shift

The graph of f ( x ) = x 2 + k f ( x ) = x 2 + k shifts the graph of f ( x ) = x 2 f ( x ) = x 2 vertically k units.

• If k > 0, shift the parabola vertically up k units.
• If k < 0, shift the parabola vertically down | k | | k | units.

Now that we have seen the effect of the constant, k , it is easy to graph functions of the form f ( x ) = x 2 + k . f ( x ) = x 2 + k . We just start with the basic parabola of f ( x ) = x 2 f ( x ) = x 2 and then shift it up or down.

It may be helpful to practice sketching f ( x ) = x 2 f ( x ) = x 2 quickly. We know the values and can sketch the graph from there.

Once we know this parabola, it will be easy to apply the transformations. The next example will require a vertical shift.

Example 9.54

Graph f ( x ) = x 2 − 3 f ( x ) = x 2 − 3 using a vertical shift.

Try It 9.107

Graph f ( x ) = x 2 − 5 f ( x ) = x 2 − 5 using a vertical shift.

Try It 9.108

Graph f ( x ) = x 2 + 7 f ( x ) = x 2 + 7 using a vertical shift.

Graph Quadratic Functions of the form f ( x ) = ( x − h ) 2 f ( x ) = ( x − h ) 2

In the first example, we graphed the quadratic function f ( x ) = x 2 f ( x ) = x 2 by plotting points and then saw the effect of adding a constant k to the function had on the resulting graph of the new function f ( x ) = x 2 + k . f ( x ) = x 2 + k .

We will now explore the effect of subtracting a constant, h , from x has on the resulting graph of the new function f ( x ) = ( x − h ) 2 . f ( x ) = ( x − h ) 2 .

Example 9.55

Graph f ( x ) = x 2 , g ( x ) = ( x − 1 ) 2 , f ( x ) = x 2 , g ( x ) = ( x − 1 ) 2 , and h ( x ) = ( x + 1 ) 2 h ( x ) = ( x + 1 ) 2 on the same rectangular coordinate system. Describe what effect adding a constant to the function has on the basic parabola.

The g ( x ) values and the h ( x ) values share the common numbers 0, 1, 4, 9, and 16, but are shifted.

Try It 9.109

ⓐ Graph f ( x ) = x 2 , g ( x ) = ( x + 2 ) 2 , f ( x ) = x 2 , g ( x ) = ( x + 2 ) 2 , and h ( x ) = ( x − 2 ) 2 h ( x ) = ( x − 2 ) 2 on the same rectangular coordinate system. ⓑ Describe what effect adding a constant to the function has on the basic parabola.

Try It 9.110

ⓐ Graph f ( x ) = x 2 , g ( x ) = x 2 + 5 , f ( x ) = x 2 , g ( x ) = x 2 + 5 , and h ( x ) = x 2 − 5 h ( x ) = x 2 − 5 on the same rectangular coordinate system. ⓑ Describe what effect adding a constant to the function has on the basic parabola.

The last example shows us that to graph a quadratic function of the form f ( x ) = ( x − h ) 2 , f ( x ) = ( x − h ) 2 , we take the basic parabola graph of f ( x ) = x 2 f ( x ) = x 2 and shift it left ( h > 0) or shift it right ( h < 0).

This transformation is called a horizontal shift .

Graph a Quadratic Function of the form f ( x ) = ( x − h ) 2 f ( x ) = ( x − h ) 2 Using a Horizontal Shift

The graph of f ( x ) = ( x − h ) 2 f ( x ) = ( x − h ) 2 shifts the graph of f ( x ) = x 2 f ( x ) = x 2 horizontally h h units.

• If h > 0, shift the parabola horizontally left h units.
• If h < 0, shift the parabola horizontally right | h | | h | units.

Now that we have seen the effect of the constant, h , it is easy to graph functions of the form f ( x ) = ( x − h ) 2 . f ( x ) = ( x − h ) 2 . We just start with the basic parabola of f ( x ) = x 2 f ( x ) = x 2 and then shift it left or right.

The next example will require a horizontal shift.

Example 9.56

Graph f ( x ) = ( x − 6 ) 2 f ( x ) = ( x − 6 ) 2 using a horizontal shift.

Try It 9.111

Graph f ( x ) = ( x − 4 ) 2 f ( x ) = ( x − 4 ) 2 using a horizontal shift.

Try It 9.112

Graph f ( x ) = ( x + 6 ) 2 f ( x ) = ( x + 6 ) 2 using a horizontal shift.

Now that we know the effect of the constants h and k , we will graph a quadratic function of the form f ( x ) = ( x − h ) 2 + k f ( x ) = ( x − h ) 2 + k by first drawing the basic parabola and then making a horizontal shift followed by a vertical shift. We could do the vertical shift followed by the horizontal shift, but most students prefer the horizontal shift followed by the vertical.

Example 9.57

Graph f ( x ) = ( x + 1 ) 2 − 2 f ( x ) = ( x + 1 ) 2 − 2 using transformations.

This function will involve two transformations and we need a plan.

Let’s first identify the constants h , k .

The h constant gives us a horizontal shift and the k gives us a vertical shift.

We first draw the graph of f ( x ) = x 2 f ( x ) = x 2 on the grid.

Try It 9.113

Graph f ( x ) = ( x + 2 ) 2 − 3 f ( x ) = ( x + 2 ) 2 − 3 using transformations.

Try It 9.114

Graph f ( x ) = ( x − 3 ) 2 + 1 f ( x ) = ( x − 3 ) 2 + 1 using transformations.

Graph Quadratic Functions of the Form f ( x ) = a x 2 f ( x ) = a x 2

So far we graphed the quadratic function f ( x ) = x 2 f ( x ) = x 2 and then saw the effect of including a constant h or k in the equation had on the resulting graph of the new function. We will now explore the effect of the coefficient a on the resulting graph of the new function f ( x ) = a x 2 . f ( x ) = a x 2 .

If we graph these functions, we can see the effect of the constant a , assuming a > 0.

To graph a function with constant a it is easiest to choose a few points on f ( x ) = x 2 f ( x ) = x 2 and multiply the y -values by a .

Graph of a Quadratic Function of the form f ( x ) = a x 2 f ( x ) = a x 2

The coefficient a in the function f ( x ) = a x 2 f ( x ) = a x 2 affects the graph of f ( x ) = x 2 f ( x ) = x 2 by stretching or compressing it.

• If 0 < | a | < 1 , 0 < | a | < 1 , the graph of f ( x ) = a x 2 f ( x ) = a x 2 will be “wider” than the graph of f ( x ) = x 2 . f ( x ) = x 2 .
• If | a | > 1 | a | > 1 , the graph of f ( x ) = a x 2 f ( x ) = a x 2 will be “skinnier” than the graph of f ( x ) = x 2 . f ( x ) = x 2 .

Example 9.58

Graph f ( x ) = 3 x 2 . f ( x ) = 3 x 2 .

We will graph the functions f ( x ) = x 2 f ( x ) = x 2 and g ( x ) = 3 x 2 g ( x ) = 3 x 2 on the same grid. We will choose a few points on f ( x ) = x 2 f ( x ) = x 2 and then multiply the y -values by 3 to get the points for g ( x ) = 3 x 2 . g ( x ) = 3 x 2 .

Try It 9.115

Graph f ( x ) = −3 x 2 . f ( x ) = −3 x 2 .

Try It 9.116

Graph f ( x ) = 2 x 2 . f ( x ) = 2 x 2 .

Graph Quadratic Functions Using Transformations

We have learned how the constants a , h , and k in the functions, f ( x ) = x 2 + k , f ( x ) = ( x − h ) 2 , f ( x ) = x 2 + k , f ( x ) = ( x − h ) 2 , and f ( x ) = a x 2 f ( x ) = a x 2 affect their graphs. We can now put this together and graph quadratic functions f ( x ) = a x 2 + b x + c f ( x ) = a x 2 + b x + c by first putting them into the form f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k by completing the square. This form is sometimes known as the vertex form or standard form.

We must be careful to both add and subtract the number to the SAME side of the function to complete the square. We cannot add the number to both sides as we did when we completed the square with quadratic equations.

When we complete the square in a function with a coefficient of x 2 that is not one, we have to factor that coefficient from just the x -terms. We do not factor it from the constant term. It is often helpful to move the constant term a bit to the right to make it easier to focus only on the x -terms.

Once we get the constant we want to complete the square, we must remember to multiply it by that coefficient before we then subtract it.

Example 9.59

Rewrite f ( x ) = −3 x 2 − 6 x − 1 f ( x ) = −3 x 2 − 6 x − 1 in the f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form by completing the square.

Try It 9.117

Rewrite f ( x ) = −4 x 2 − 8 x + 1 f ( x ) = −4 x 2 − 8 x + 1 in the f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form by completing the square.

Try It 9.118

Rewrite f ( x ) = 2 x 2 − 8 x + 3 f ( x ) = 2 x 2 − 8 x + 3 in the f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form by completing the square.

Once we put the function into the f ( x ) = ( x − h ) 2 + k f ( x ) = ( x − h ) 2 + k form, we can then use the transformations as we did in the last few problems. The next example will show us how to do this.

Example 9.60

Graph f ( x ) = x 2 + 6 x + 5 f ( x ) = x 2 + 6 x + 5 by using transformations.

Step 1. Rewrite the function in f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k vertex form by completing the square.

Step 2: Graph the function using transformations.

Looking at the h , k values, we see the graph will take the graph of f ( x ) = x 2 f ( x ) = x 2 and shift it to the left 3 units and down 4 units.

Try It 9.119

Graph f ( x ) = x 2 + 2 x − 3 f ( x ) = x 2 + 2 x − 3 by using transformations.

Try It 9.120

Graph f ( x ) = x 2 − 8 x + 12 f ( x ) = x 2 − 8 x + 12 by using transformations.

We list the steps to take to graph a quadratic function using transformations here.

Graph a quadratic function using transformations.

• Step 1. Rewrite the function in f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form by completing the square.

Step 2. Graph the function using transformations.

Example 9.61

Graph f ( x ) = −2 x 2 − 4 x + 2 f ( x ) = −2 x 2 − 4 x + 2 by using transformations.

Try It 9.121

Graph f ( x ) = −3 x 2 + 12 x − 4 f ( x ) = −3 x 2 + 12 x − 4 by using transformations.

Try It 9.122

Graph f ( x ) = −2 x 2 + 12 x − 9 f ( x ) = −2 x 2 + 12 x − 9 by using transformations.

Now that we have completed the square to put a quadratic function into f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form, we can also use this technique to graph the function using its properties as in the previous section.

If we look back at the last few examples, we see that the vertex is related to the constants h and k .

In each case, the vertex is ( h , k ). Also the axis of symmetry is the line x = h .

We rewrite our steps for graphing a quadratic function using properties for when the function is in f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form.

Graph a quadratic function in the form f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k using properties.

• Step 1. Rewrite the function in f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form.
• Step 2. Determine whether the parabola opens upward, a > 0, or downward, a < 0.
• Step 3. Find the axis of symmetry, x = h .
• Step 4. Find the vertex, ( h , k ).
• Step 5. Find the y -intercept. Find the point symmetric to the y -intercept across the axis of symmetry.
• Step 6. Find the x -intercepts.
• Step 7. Graph the parabola.

Example 9.62

ⓐ Rewrite f ( x ) = 2 x 2 + 4 x + 5 f ( x ) = 2 x 2 + 4 x + 5 in f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form and ⓑ graph the function using properties.

Try It 9.123

ⓐ Rewrite f ( x ) = 3 x 2 − 6 x + 5 f ( x ) = 3 x 2 − 6 x + 5 in f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form and ⓑ graph the function using properties.

Try It 9.124

ⓐ Rewrite f ( x ) = −2 x 2 + 8 x − 7 f ( x ) = −2 x 2 + 8 x − 7 in f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form and ⓑ graph the function using properties.

Find a Quadratic Function from its Graph

So far we have started with a function and then found its graph.

Now we are going to reverse the process. Starting with the graph, we will find the function.

Example 9.63

Determine the quadratic function whose graph is shown.

Since it is quadratic, we start with the f ( x ) = a ( x − h ) 2 + k form. The vertex, ( h , k ) , is ( −2 , −1 ) so h = −2 and k = −1. f ( x ) = a ( x − ( −2 ) ) 2 − 1 To find a , we use the y -intercept, ( 0 , 7 ) . So f ( 0 ) = 7 . 7 = a ( 0 + 2 ) 2 − 1 Solve for a . 7 = 4 a − 1 8 = 4 a 2 = a Write the function. f ( x ) = a ( x − h ) 2 + k Substitute in h = −2 , k = −1 and a = 2 . f ( x ) = 2 ( x + 2 ) 2 − 1 Since it is quadratic, we start with the f ( x ) = a ( x − h ) 2 + k form. The vertex, ( h , k ) , is ( −2 , −1 ) so h = −2 and k = −1. f ( x ) = a ( x − ( −2 ) ) 2 − 1 To find a , we use the y -intercept, ( 0 , 7 ) . So f ( 0 ) = 7 . 7 = a ( 0 + 2 ) 2 − 1 Solve for a . 7 = 4 a − 1 8 = 4 a 2 = a Write the function. f ( x ) = a ( x − h ) 2 + k Substitute in h = −2 , k = −1 and a = 2 . f ( x ) = 2 ( x + 2 ) 2 − 1

Try It 9.125

Write the quadratic function in f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form whose graph is shown.

Try It 9.126

Access these online resources for additional instruction and practice with graphing quadratic functions using transformations.

• Function Shift Rules Applied to Quadratic Functions
• Changing a Quadratic from Standard Form to Vertex Form
• Using Transformations to Graph Quadratic Functions
• Finding Quadratic Equation in Vertex Form from Graph

Section 9.7 Exercises

Practice makes perfect.

In the following exercises, ⓐ graph the quadratic functions on the same rectangular coordinate system and ⓑ describe what effect adding a constant, k , to the function has on the basic parabola.

f ( x ) = x 2 , g ( x ) = x 2 + 4 , f ( x ) = x 2 , g ( x ) = x 2 + 4 , and h ( x ) = x 2 − 4 . h ( x ) = x 2 − 4 .

f ( x ) = x 2 , g ( x ) = x 2 + 7 , f ( x ) = x 2 , g ( x ) = x 2 + 7 , and h ( x ) = x 2 − 7 . h ( x ) = x 2 − 7 .

In the following exercises, graph each function using a vertical shift.

f ( x ) = x 2 + 3 f ( x ) = x 2 + 3

f ( x ) = x 2 − 7 f ( x ) = x 2 − 7

g ( x ) = x 2 + 2 g ( x ) = x 2 + 2

g ( x ) = x 2 + 5 g ( x ) = x 2 + 5

h ( x ) = x 2 − 4 h ( x ) = x 2 − 4

h ( x ) = x 2 − 5 h ( x ) = x 2 − 5

In the following exercises, ⓐ graph the quadratic functions on the same rectangular coordinate system and ⓑ describe what effect adding a constant, h h , inside the parentheses has

f ( x ) = x 2 , g ( x ) = ( x − 3 ) 2 , f ( x ) = x 2 , g ( x ) = ( x − 3 ) 2 , and h ( x ) = ( x + 3 ) 2 . h ( x ) = ( x + 3 ) 2 .

f ( x ) = x 2 , g ( x ) = ( x + 4 ) 2 , f ( x ) = x 2 , g ( x ) = ( x + 4 ) 2 , and h ( x ) = ( x − 4 ) 2 . h ( x ) = ( x − 4 ) 2 .

In the following exercises, graph each function using a horizontal shift.

f ( x ) = ( x − 2 ) 2 f ( x ) = ( x − 2 ) 2

f ( x ) = ( x − 1 ) 2 f ( x ) = ( x − 1 ) 2

f ( x ) = ( x + 5 ) 2 f ( x ) = ( x + 5 ) 2

f ( x ) = ( x + 3 ) 2 f ( x ) = ( x + 3 ) 2

f ( x ) = ( x − 5 ) 2 f ( x ) = ( x − 5 ) 2

f ( x ) = ( x + 2 ) 2 f ( x ) = ( x + 2 ) 2

In the following exercises, graph each function using transformations.

f ( x ) = ( x + 2 ) 2 + 1 f ( x ) = ( x + 2 ) 2 + 1

f ( x ) = ( x + 4 ) 2 + 2 f ( x ) = ( x + 4 ) 2 + 2

f ( x ) = ( x − 1 ) 2 + 5 f ( x ) = ( x − 1 ) 2 + 5

f ( x ) = ( x − 3 ) 2 + 4 f ( x ) = ( x − 3 ) 2 + 4

f ( x ) = ( x + 3 ) 2 − 1 f ( x ) = ( x + 3 ) 2 − 1

f ( x ) = ( x + 5 ) 2 − 2 f ( x ) = ( x + 5 ) 2 − 2

f ( x ) = ( x − 4 ) 2 − 3 f ( x ) = ( x − 4 ) 2 − 3

f ( x ) = ( x − 6 ) 2 − 2 f ( x ) = ( x − 6 ) 2 − 2

Graph Quadratic Functions of the form f ( x ) = a x 2 f ( x ) = a x 2

In the following exercises, graph each function.

f ( x ) = −2 x 2 f ( x ) = −2 x 2

f ( x ) = 4 x 2 f ( x ) = 4 x 2

f ( x ) = −4 x 2 f ( x ) = −4 x 2

f ( x ) = − x 2 f ( x ) = − x 2

f ( x ) = 1 2 x 2 f ( x ) = 1 2 x 2

f ( x ) = 1 3 x 2 f ( x ) = 1 3 x 2

f ( x ) = 1 4 x 2 f ( x ) = 1 4 x 2

f ( x ) = − 1 2 x 2 f ( x ) = − 1 2 x 2

In the following exercises, rewrite each function in the f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form by completing the square.

f ( x ) = −3 x 2 − 12 x − 5 f ( x ) = −3 x 2 − 12 x − 5

f ( x ) = 2 x 2 − 12 x + 7 f ( x ) = 2 x 2 − 12 x + 7

f ( x ) = 3 x 2 + 6 x − 1 f ( x ) = 3 x 2 + 6 x − 1

f ( x ) = −4 x 2 − 16 x − 9 f ( x ) = −4 x 2 − 16 x − 9

In the following exercises, ⓐ rewrite each function in f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form and ⓑ graph it by using transformations.

f ( x ) = x 2 + 6 x + 5 f ( x ) = x 2 + 6 x + 5

f ( x ) = x 2 + 4 x − 12 f ( x ) = x 2 + 4 x − 12

f ( x ) = x 2 + 4 x + 3 f ( x ) = x 2 + 4 x + 3

f ( x ) = x 2 − 6 x + 8 f ( x ) = x 2 − 6 x + 8

f ( x ) = x 2 − 6 x + 15 f ( x ) = x 2 − 6 x + 15

f ( x ) = x 2 + 8 x + 10 f ( x ) = x 2 + 8 x + 10

f ( x ) = − x 2 + 8 x − 16 f ( x ) = − x 2 + 8 x − 16

f ( x ) = − x 2 + 2 x − 7 f ( x ) = − x 2 + 2 x − 7

f ( x ) = − x 2 − 4 x + 2 f ( x ) = − x 2 − 4 x + 2

f ( x ) = − x 2 + 4 x − 5 f ( x ) = − x 2 + 4 x − 5

f ( x ) = 5 x 2 − 10 x + 8 f ( x ) = 5 x 2 − 10 x + 8

f ( x ) = 3 x 2 + 18 x + 20 f ( x ) = 3 x 2 + 18 x + 20

f ( x ) = 2 x 2 − 4 x + 1 f ( x ) = 2 x 2 − 4 x + 1

f ( x ) = 3 x 2 − 6 x − 1 f ( x ) = 3 x 2 − 6 x − 1

f ( x ) = −2 x 2 + 8 x − 10 f ( x ) = −2 x 2 + 8 x − 10

f ( x ) = −3 x 2 + 6 x + 1 f ( x ) = −3 x 2 + 6 x + 1

In the following exercises, ⓐ rewrite each function in f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form and ⓑ graph it using properties.

f ( x ) = 2 x 2 + 4 x + 6 f ( x ) = 2 x 2 + 4 x + 6

f ( x ) = 3 x 2 − 12 x + 7 f ( x ) = 3 x 2 − 12 x + 7

f ( x ) = − x 2 + 2 x − 4 f ( x ) = − x 2 + 2 x − 4

f ( x ) = −2 x 2 − 4 x − 5 f ( x ) = −2 x 2 − 4 x − 5

In the following exercises, match the graphs to one of the following functions: ⓐ f ( x ) = x 2 + 4 f ( x ) = x 2 + 4 ⓑ f ( x ) = x 2 − 4 f ( x ) = x 2 − 4 ⓒ f ( x ) = ( x + 4 ) 2 f ( x ) = ( x + 4 ) 2 ⓓ f ( x ) = ( x − 4 ) 2 f ( x ) = ( x − 4 ) 2 ⓔ f ( x ) = ( x + 4 ) 2 − 4 f ( x ) = ( x + 4 ) 2 − 4 ⓕ f ( x ) = ( x + 4 ) 2 + 4 f ( x ) = ( x + 4 ) 2 + 4 ⓖ f ( x ) = ( x − 4 ) 2 − 4 f ( x ) = ( x − 4 ) 2 − 4 ⓗ f ( x ) = ( x − 4 ) 2 + 4 f ( x ) = ( x − 4 ) 2 + 4

In the following exercises, write the quadratic function in f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k form whose graph is shown.

Writing Exercise

Graph the quadratic function f ( x ) = x 2 + 4 x + 5 f ( x ) = x 2 + 4 x + 5 first using the properties as we did in the last section and then graph it using transformations. Which method do you prefer? Why?

Graph the quadratic function f ( x ) = 2 x 2 − 4 x − 3 f ( x ) = 2 x 2 − 4 x − 3 first using the properties as we did in the last section and then graph it using transformations. Which method do you prefer? Why?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

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Access for free at https://openstax.org/books/intermediate-algebra/pages/1-introduction

• Authors: Lynn Marecek
• Publisher/website: OpenStax
• Book title: Intermediate Algebra
• Publication date: Mar 14, 2017
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra/pages/9-7-graph-quadratic-functions-using-transformations

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