Graph Quadratic Functions Using Transformations

9.7 Graph Quadratic Functions Using Transformations

The topics covered in this section are:

  1. Graph quadratic equations of the form $f(x)=x^{2}+k$
  2. Graph quadratic functions of the form $f(x)=(x-h)^{2}$
  3. Graph quadratic functions of the form $f(x)=ax^{2}$
  4. Graph quadratic functions using transformations
  5. Find a quadratic function from its graph

9.7.1 Graph Quadratic Functions of the form $f(x)=x^{2}+k$

In the last section, we learned how to graph quadratic functions using their properties. Another method involves starting with the basic graph of $f(x)=x^{2}$ 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 $f(x)=x^{2}$ 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 $f(x)=x^{2}+k$.

Example 1

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

Solution

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

A table depicting the effect of constants on the basic function of x squared. The table has seven columns labeled x, f of x equals x squared, the ordered pair (x, f of x), g of x equals x squared plus 2, the ordered pair (x, g of x), h of x equals x squared minus 2, and the ordered pair (x, h of x). In the x column, the values given are negative 3, negative 2, negative 1, 0, 1, 2, and 3. In the f of x equals x squared column, the values are 9, 4, 1, 0, 1, 4, and 9. In the (x, f of x) column, the ordered pairs (negative 3, 9), (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), (2, 4), and (3, 9) are given. The g of x equals x squared plus 2 column contains the expressions 9 plus 2, 4 plus 2, 1 plus 2, 0 plus 2, 1 plus 2, 4 plus 2, and 9 plus 2. The (x, g of x) column has the ordered pairs of (negative 3, 11), (negative 2, 6), (negative 1, 3), (0, 2), (1, 3), (2, 6), and (3, 11). In the h of x equals x squared minus 2 column, the expressions given are 9 minus 2, 4 minus 2, 1 minus 2, 0 minus 2, 1 minus 2, 4 minus 2, and 9 minus 2. In last column, (x, h of x), contains the ordered pairs (negative 3, 7), (negative 2, 2), (negative 1, negative 1), (0, negative 2), (1, negative 1), (2, 2), and (3, 7).

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.

This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle is the graph of f of x equals x squared has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top parabola has been moved up 2 units, and the bottom has been moved down 2 units.

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

The graph of $h(x)=x^{2}+2$ is the same as the graph of $f(x)=x^{2}$ but shifted down $2$ units.

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

This transformation is called a vertical shift.

GRAPH A QUADRATIC FUNCTION OF THE FORM $f(x)=x^{2}$ USING A VERTICAL SHIFT

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

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

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

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

This figure shows an upward-opening parabola on the x y-coordinate plane, with vertex (0, 0). Other points on the curve are located at (negative 4, 16), (negative 3, 9), (negative 2, 4), (negative 1, 1), (1, 1), (2, 4), (3, 9), and (4, 16).

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

Example 2

Graph $f(x)=x^{2}-3$ using a vertical shift.

Solution

We first draw the graph of $f(x) =x^{2}$ on the grid.This figure shows an upward-opening parabola on the x y-coordinate plane with a vertex of (0, 0) with other points on the curve located at (negative 1, 1) and (1, 1). It is the graph of f of x equals x squared.
Determine $k$.$f(x)=x^{2}+k$
$f(x)=x^{2}-3$
$k=-3$
Shift the graph $f(x)=x^{2}$ down $3$.This figure shows 2 upward-opening parabolas on the x y-coordinate plane. The top curve is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The bottom curve has been moved down 3 units.

9.7.2 Graph Quadratic Functions of the form $f(x)=(x-h)^{2}$

In the first example, we graphed the quadratic function $f(x)=x^{2}$ 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 $f(x)=x^{2}+k$.

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

Example 3

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

Solution

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

A table depicting the effect of constants on the basic function of x squared. The table has seven columns labeled x, f of x equals x squared, the ordered pair (x, f of x), g of x equals the quantity of x minus 1 squared, the ordered pair (x, g of x), h of x equals the quantity of x plus 1 squared, and the ordered pair (x, h of x). In the x column, the values given are negative 3, negative 2, negative 1, 0, 1, 2, and 3. In the f of x equals x squared column, the values are 9, 4, 1, 0, 1, 4, and 9. In the (x, f of x) column, the ordered pairs (negative 3, 9), (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), (2, 4), and (3, 9) are given. The g of x equals the quantity of x minus 1 squared column contains the values of 16, 9, 4, 1, 0, 1, and 4. The (x, g of x) column has the ordered pairs of (negative 3, 1), (negative 2, 9), (negative 1, 4), (0, 1), (1, 0), (2, 1), and (3, 4). In the h of x equals the quantity of x plus 1 squared, the values given are 4, 1, 0, 1, 4, 9, and 16. In last column, (x, h of x), contains the ordered pairs (negative 3, 4), (negative 2, 1), (negative 1, 0), (0, 4), (1, negative 1), (2, 9), and (3, 16).

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

This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The left curve has been moved to the left 1 unit, and the right curve has been moved to the right 1 unit.
The figure says on the first line that the graph of g of x equals the quantity x minus 1 square is the same as the graph of f of x equals x squared but shifted right 1 unit. The second line states that the graph of h of x equals the quantity x plus 1 squared is the same as the graph of f of x equals x squared but shifted left 1 unit. The third line of the figure says g of x equals the quantity x minus 1 squared with an arrow underneath it pointing to the right with 1 unit written beside it. Finally, it gives h of x equals the quantity of x plus 1 squared with an arrow underneath it pointing to the left with 1 unit written beside it.

The last example shows us that to graph a quadratic function of the form $f(x)=(x-h)^{2}$, we take the basic parabola graph of $f(x)=x^{2}$ 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 $f(x)=(x-h)^{2}$ USING A HORIZONTAL SHIFT

The graph of $f(x)=(x-h)^{2}$ shifts the graph of $f(x)=x^{2}$ horizontally $h$ units.

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

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

The next example will require a horizontal shift.

Example 4

Graph $f(x)=(x-6)^{2}$ using a horizontal shift.

Solution

We first draw the graph of $f(x)=x^{2}$ on the grid..
Determine $h$.$f(x)=(x-h)^{2}$
$f(x)=(x-6)^{2}$
$h=6$
Shift the graph $f(x)=x^{2}$ to the right $6$ units..

Now that we know the effect of the constants h and k, we will graph a quadratic function of the form $f(x)=(x-h)^{2} +k$ 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 5

Graph $f(x)=(x+1)^{2}-2$ using transformations.

Solution

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

Let’s first identify the constants $h,k$.

$f(x)=(x+1)^{2}-2$

$f(x)=(x-h)^{2}+k$

$f(x)=(x-(-1))^{2}+(-2)$

$h=-1$ and $k=-2$

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

F of x equals x squared is given with an arrow coming from it pointing to f of x equals the quantity x plus 1 squared with an arrow coming from it pointing to f of x equals the quantity x plus 1 squared minus 2. The next lines say h equals negative 1 which means shift left 1 unit and k equals negative 2 which means shift down 2 units.

We first draw the graph of $f(x)=x^{2}$ on the grid.

The figure says on the first line that the graph of f of x equals the quantity x plus 1 squared is the same as the graph of f of x equals x squared but shifted left 1 unit. The second line states that the graph of f of x equals the quantity x plus 1 squared minus 2 is the same as the graph of f of x equals the quantity x plus 1 squared but shifted down 2 units.
The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By shifting that graph of f of x equals x squared left 1, we move to the next graph, which shows the original f of x equals x squared and then another curve moved left one unit to produce f of x equals the quantity of x plus 1 squared. By moving f of x equals the quantity of x plus 1 squared down 1, we move to the final graph, which shows the original f of x equals x squared and the f of x equals the quantity of x plus 1, then another curve moved down 1 to produce f of x equals the quantity of x plus 1 squared minus 2.

9.7.3 Graph Quadratic Functions of the Form $f(x)=ax^{2}$

So far we graphed the quadratic function $f(x)=x^{2}$ 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 $f(x)=ax^{2}$.

A table depicting the effect of constants on the basic function of x squared. The table has seven columns labeled x, f of x equals x squared, the ordered pair (x, f of x), g of x equals 2 times x squared, the ordered pair (x, g of x), h of x equals one-half times x squared, and the ordered pair (x, h of x). In the x column, the values given are negative 2, negative 1, 0, 1, and 2. In the f of x equals x squared column, the values are 4, 1, 0, 1, and 4. In the (x, f of x) column, the ordered pairs (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4) are given. The g of x equals 2 times x squared column contains the expressions 2 times 4, 2 times 1, 2 times 0, 2 times 1, and 2 times 4. The (x, g of x) column has the ordered pairs of (negative 2, 8), (negative 1, 2), (0, 0), (1, 2), and (2,8). In the h of x equals one-half times x squared, the expressions given are one-half times 4, one-half times 1, one-half times 0, one-half times 1, and one-half times 4. In last column, (x, h of x), contains the ordered pairs (negative 2, 2), (negative 1, one-half), (0, 0), (1, one-half), and (2, 2).

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

This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The slimmer curve of g of x equals 2 times x square has a vertex at (0,0) and other points of (negative 1, one-half) and (1, one-half). The wider curve, h of x equals one-half x squared, has a vertex at (0,0) and other points of (negative 2, 2) and (2,2).

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

GRAPH OF A QUADRATIC FUNCTION OF THE FORM $f(x)=ax^{2}$

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

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

Example 6

Graph $f(x)=3x^{2}$.

Solution

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

The table depicts the effect of constants on the basic function of x squared. The table has 3 columns labeled x, f of x equals x squared with the ordered pair (x, f of x), and g of x equals 3 times x squared with the ordered pair (x, g of x). In the x column, the values given are negative 2, negative 1, 0, 1, and 2. In the f of x equals x squared with the ordered pair (x, f of x), the ordered pairs (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4) are given. The g of x equals 3 times x squared with the ordered pair (x, g of x) column has the ordered pairs of (negative 2, 12) because 3 times 4 equals 12, (negative 1, 3) because 3 times 1 equals 3, (0, 0) because 3 times 0 equals 0, (1, 3) because 3 times 1 equals 3, and (2,12) because 3 times 4 equals 12. The graph beside the table shows 2 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points given on the curve are located at (negative 2, 4) (negative 1, 1), (1, 1), and (2,4). The slimmer curve of g of x equals 3 times x squared has a vertex at (0,0) and other points given of (negative 2, 12), (negative 1, 3), (1, 3), and (2,12).

9.7.4 Graph Quadratic Functions Using Transformations

We have learned how the constants $a,h,$ and $k$ in the functions, $f(x)=x^{2}, f(x)=(x-h)^{2},$ and $f(x)=ax^{2}$ affect their graphs. We can now put this together and graph quadratic functions $f(x)=ax^{2}+bx+c$ by first putting them into the form $f(x)=a(x-h)^{2}+k$ 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.

This figure shows the difference when completing the square with a quadratic equation and a quadratic function. For the quadratic equation, start with x squared plus 8 times x plus 6 equals zero. Subtract 6 from both sides to get x squared plus 8 times x equals negative 6 while leaving space to complete the square. Then, complete the square by adding 16 to both sides to get x squared plush 8 times x plush 16 equals negative 6 plush 16. Factor to get the quantity x plus 4 squared equals 10. For the quadratic function, start with f of x equals x squared plus 8 times x plus 6. The second line shows to leave space between the 8 times x and the 6 in order to complete the square. Complete the square by adding 16 and subtracting 16 on the same side to get f of x equals x squared plus 8 times x plush 16 plus 6 minus 16. Factor to get f of x equals the quantity of x plush 4 squared minus 10.

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 7

Rewrite $f(x)=-3x^{2}-6x-1$ in the $f(x)=a(x-h)^{2}+k$ form by completing the square.

Solution

$f(x)=-3x^{+2}-6x-1$
Separate the $x$ terms from the constant.$f(x)=-3x^{+2}-6x \ \ \ \ -1$
Factor the coefficient of $x^{2}, -3$$f(x)=-3(x^{2}+2x) \ \ \ \ -1$
Prepare to complete the square.$f(x)=-3(x^{2}+2x \ \ \ \ )-1$
Take half of $2$ and the square it to complete the square. $(\frac{1}{2} \cdot 2)^{2} =1$
The constant $1$ completes the square in the parentheses, but the parentheses is multiplied by $-3$. So we are really adding $-3$. We must then add$3$ to not change the value of the function..
Rewrite the trinomial as a square and subtract the constants.$f(x)=-3(x+1)+2$
The function is now in the $f(x)=a(x-h)^{2}+k$ form.$f(x)=a(x-h)^{2} +k$
$f(x)=-3(x+1)^{2}+2$

Once we put the function into the $f(x)=(x-h)^{2}+k$ 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 8

Graph $f(x)=x^{2}+6x+5$ by using transofmraions.

Solution

Step 1. Rewrite the function in $f(x)=a(x-h)^{2}+k$ vertex form by completing the square.

$f(x)=x^{2}+6x+5$
Seperate the $x$ terms from the constant.$f(x)=x^{2}+6x+5$
Take half of $6$ and the square it to complete the square.
$(\frac{1}{2} \cdot 6)^{2} =9$
We both add $9$ and subtract $9$ to not change the value of the function$$f(x)=x^{2}+6x+9+5-9$
Rewrite the trinomial as a square and subtract the constants.$f(x)=(x+3)^{2}-4$
The function is now in the $f(x)=(x-h)^{2}+k$ form.$f(x)=(x-h)^{2}+k$
$f(x)=(x+3)^{2}-4$

Step 2. Graph the function using transformations.

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

F of x equals x squared is given with an arrow coming from it pointing to f of x equals the quantity x plus 3 squared with an arrow coming from it pointing to f of x equals the quantity x plus 3 squared minus 4. The next lines say h equals negative 3 which means shift left 3 unit and k equals negative 4 which means shift down 4 units

We first draw the graph of $f(x)=x^{2}$ on the grid.

To graph f of x equals the quantity x plus 3 squared, shift the graph of f of x equals x squares to the left 3 units. To graph f of x equals the quantity x plus 3 squared minus 4, shift the graph the quantity x plus 3 squared down 4 units.
The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By shifting that graph of f of x equals x squared left 3, we move to the next graph, which shows the original f of x equals x squared and then another curve moved left 3 units to produce f of x equals the quantity of x plus 3 squared. By moving f of x equals the quantity of x plus 3 squared down 2, we move to the final graph, which shows the original f of x equals x squared and the f of x equals the quantity of x plus 3 squared, then another curve moved down 4 to produce f of x equals the quantity of x plus 1 squared minus 4.

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

HOW TO: Graph a quadratic function using transformations.

  1. Rewrite the function in $f(x)=a(x-h)^{2}+k$ form by completing the square.
  2. Graph the function using transformations.

Example 9

Graph $f(x)=-2x^{2}-4x+2$ by using transformations.

Solution

Step 1. Rewrite the function in $f(x)=a(x-h)^{2}+k$ vertex form by completing the square.

$f(x)=-2x^{2}-4x+2$
Separate the $x$ terms from the constant.$f(x)=-2x^{2}-4x+2$
We need the coefficient of $x^{2}$ to be one.
We factor $-2$ from the $x$-terms.
$f(x)=-2(x^{2}+2x)+2$
Take half of $2$ and then square it to complete the square.
$(\frac{1}{2} \cdot 2)^{2}=1$
We add $1$ to complete the square in the parentheses, bu the parentheses is multiplied by $-2$. So we are really adding $-2$. To not change the value of the function we add $2$.$f(x)=-2(x^{2}+2x+1)+2+2$
Rewrite the trinomial as a square and subtract the constants.$f(x)=-2(x+1)^{2}+4$
The function is now in the $f(x)=a(x-h)^{2}+k$ form.$f(x)=a(x-h)^{2}+k$
$f(x)=-2(x+1)^{2}+4$

Step 2. Graph the function using transformations.

F of x equals x squared is given with an arrow coming from it pointing to f of x equals negative 2 times x squared with an arrow coming from it pointing to f of x equals negative 2 times the quantity x plus 1 squared. An arrow come from it to point to f of x equals negative 2 times the quantity x plus 1 squared plus 4. The next line says a equals negative 2 which means multiply the y-values by negative 2, then h equals negative 1 which means shift left 1 unit and k equals 4 which means shift up 4 units

We first draw the graph of $f(x)=x^{2}$ on the grid.

To graph f of x equals negative 2 times x squared, multiply the y-values in parabola of f of x equals x squared by negative 2. To graph f of x equals negative 2 times the quantity x plus 1 squared, shift the graph of f of x equals negative 2 times x squared to the left 1 unit. To graph f of x equals negative 2 times the quantity x plus 1 squared plus 4, shift the graph of f of x equals negative 2 times the quantity x plus 1 squared up 4 units.
The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By multiplying by negative 2, move to the next graph showing the original f of x equals x squared and the new slimmer and flipped graph of f of x equals negative 2 x squared. By shifting that graph of f of x equals negative 2 times x squared left 1, we move to the next graph, which shows the original f of x equals x squared, f of x equals negative 2 x squared, and then another curve moved left 1 unit to produce f of x equals negative 2 times the quantity of x plus 1 squared. By moving f of x equals negative 2 times the quantity of x plus 1 squared up 4, we move to the final graph, which shows the original f of x equals x squared, f of x equals negative 2 x squared, and the f of x equals negative 2 times the quantity of x plus 1 squared, then another curve moved up 4 to produce f of x equals negative 2 times the quantity of x plus 1 squared plus 4.

Now that we have completed the square to put a quadratic function into $f(x)=a(x-h)^{2}+k$ 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 $k$.

The first graph shows an upward-opening parabola on the x y-coordinate plane with a vertex of (negative 3, negative 4) with other points of (0, negative 5) and (0, negative 1). Underneath the graph, it shows the standard form of a parabola, f of x equals the quantity x minus h squared plus k, with the equation of the parabola f of x equals the quantity of x plus 3 squared minus 4 where h equals negative 3 and k equals negative 4. The second graph shows a downward-opening parabola on the x y-coordinate plane with a vertex of (negative 1, 4) and other points of (0,2) and (negative 2,2). Underneath the graph, it shows the standard form of a parabola, f of x equals a times the quantity x minus h squared plus k, with the equation of the parabola f of x equals negative 2 times the quantity of x plus 1 squared plus 4 where h equals negative 1 and k equals 4.

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 $f(x)=a(x-h)^{2}+k$ form.

HOW TO: Graph a quadratic function in the form $f(x)=a(x-h)^{2}+k$ using properties.

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

Example 10

  • Rewrite $f(x)=2x^{2}+4x+5$ in $f(x)=a(x-h)^{2}+5$ form
  • Graph the function using properties.
Solution

Rewrite the function $f(x)=a(x-h)^{2}+k$ form by completing the square.$f(x)=2x^{2}+4x+5$
$f(x)=2(x^{2}+2x)+5$
$f(x)=2(x^{2}+2x+1)+5-2$
$f(x)=2(x+1)^{2}+3$
Identify the constants $a,h,k$.$a=2, h=-1, k=3$
Since $a=2$, the parabola opens upward..
The axis of symmetry is $x=h$.The axis of symmetry is $x=-1$.
The vertex is $(h,k)$.The vertex is $(-1,3)$.
Find the $y$-intercept by finding $f(0)$$f(0)=2 \cdot 0^{2}+4 \cdot 0 +5$
$f(0)=5$
$y$-intercept $(0,5)$
Find the point symmetric to $(0,5)$ across the axis of symmetry.$(-2,5)$
Find the $x$-intercepts.The discriminant negative, so there are no $x$-intercepts. Graph the parabola.
.

9.7.5 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 11

Determine the quadratic function whose graph is shown.

The graph shown is an upward facing parabola with vertex (negative 2, negative 1) and y-intercept (0, 7).
Solution

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 $a$.$7=4a-1$
$8=4a$
$2=a$
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$
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