Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

Activity

When finding an exact function that has particular zeros and passes through a certain point, we use the intercept form of a quadratic function:

$f (x) = a (x - p) (x - q)$

The variables $p$ and $q$ represent the zeros or roots of the function. The variable $a$ accounts for any constant that might have been factored out when solving.

Remember, if needed in the following problems, use the "^" symbol to enter exponents.

1.

Find a function, $m (x)$ , in intercept form, that has the zeros –5 and 8.

Enter a function, $m (x)$ , in intercept form.

Compare your answer:

The intercept form is $m (x) = a (x - (- 5)) (x - 8)$ .

$= a (x + 5) (x - 8)$

Let's use the intercept form of the function above to find the exact function that passes through $(5, - 60)$ .

2.

First, find the value of $a$ . To do this, substitute the values of the point into the intercept form of the function.

Enter the value of $a$ .

Compare your answer:

$m (x) = a (x + 5) (x - 8)$

$- 60 = a ((5) + 5) ((5) - 8)$

$- 60 = a (10) (- 3)$

$- 60 = - 30 a$

$2 = a$

3.

Substitute the value of $a$ into the intercept form of the function and multiply to find the standard form.

Enter the standard form of the function.

Compare your answer:

$m (x) = a (x + 5) (x - 8)$

$m (x) = 2 (x + 5) (x - 8)$

$m (x) = 2 (x^{2} + 5 x - 8 x - 40)$

$m (x) = 2 (x^{2} - 3 x - 40)$

$m (x) = 2 x^{2} - 6 x - 80$

This is the function with zeros –5 and 8 that passes through $(5, - 60)$ .

4.

Write the quadratic function, $n (x)$ , in standard form, that has the zeros 9 and –2 and passes through the point $(0, - 36)$ .

Enter the quadratic function, $n (x)$ , in standard form.

Compare your answer:

$n (x) = 2 x^{2} - 14 x - 36$

5.

Write the quadratic function, $p (x)$ , in standard form, that has the zeros –6 and –4 and passes through the point $(- 5, 3)$ .

Enter the quadratic function, $p (x)$ , in standard form.

Compare your answer:

$p (x) = - 3 x^{2} - 30 x - 72$

6.

Write the quadratic function, $c (x)$ , in standard form, that has the zeros 1 and 7 and passes through the point $(9, 16)$ .

Enter the quadratic function, $c (x)$ , in standard form.

Compare your answer:

$c (x) = x^{2} - 8 x + 7$

Video: Finding a Quadratic Function from Its Zeros and a Point

Watch the following video to learn more about finding a quadratic function from its zeros and a point.

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Self Check

Write the quadratic function,

z(x)

, in standard form, that has the zeros –1 and 8 and passes through the point

(0, 16)

.

$z(x) = x^2 + 7x + 8$
$z(x) = 2x^2 - 14x - 16$
$z(x) = 3x^2 - 21x - 24$
$z(x) = -2x^2 + 14x + 16$

Additional Resources

Finding a Quadratic Function from Its Zeros and a Point

Think about how you would solve the quadratic equation $x^{2} - 3 x + 2 = 0$ .

The factored form is $(x - 1) (x - 2) = 0$ .

The solutions are $x = 1$ and $x = 2$ .

Now think about how you would solve $2 x^{2} - 6 x + 4 = 0$ .

Factor out the GCF, 2.

$2 (x^{2} - 3 x + 2) = 0$

It looks similar to the previous quadratic equation.

The factored form is $2 (x - 1) (x - 2) = 0$ .

Once again, the solutions are $x = 1$ and $x = 2$ .

Since they have the same solutions, these two quadratic equations are related. They have the same zeros and the same axis of symmetry.

When we are reversing this process, going from the zeros of a quadratic function to finding its standard form, it is possible there is a multiplier, such as 2 in the example above, that we do not know about.

To account for this, we use the variable $a$ as the multiplier until we find out exactly what it is.

We can incorporate $a$ into the function using the intercept form, $f (x) = a (x - p) (x - q)$ . The variables $p$ and $q$ represent the zeros or roots of the function.

Let's look at an example to help us understand.

Example 1

Write the quadratic function, $f (x)$ , that has the zeros –5 and 4 and passes through the point $(- 4, - 16)$ .

We know the zeros are –5 and 4, so the factored form of the quadratic function is:

$f (x) = (x + 5) (x - 4)$

The intercept form is $f (x) = a (x + 5) (x - 4)$ .

We are given one more piece of information that allows us to find the value of $a$ . We know that the function passes through the point $(- 4, - 16)$ .

We can substitute –4 for $x$ and –16 for the value of the function and solve for $a$ .

$f (x) = a (x + 5) (x - 4)$

$- 16 = a ((- 4) + 5) ((- 4) - 4)$

$- 16 = a (1) (- 8)$

$- 16 = - 8 a$

$2 = a$

Since we know $a = 2$ , we can substitute 2 for $a$ into the intercept form of the quadratic function.

$f (x) = a (x + 5) (x - 4)$

$f (x) = 2 (x + 5) (x - 4)$

$f (x) = 2 (x^{2} + 5 x - 4 x - 20)$

$f (x) = 2 (x^{2} + x - 20)$

$f (x) = 2 x^{2} + 2 x - 40$

The specific function that has zeros –5 and 4 and passes through the point $(- 4, - 16)$ is:

$f (x) = 2 x^{2} + 2 x - 40$

Let's graph the functions to check our work.

Two graphed parabolas, one red and one green, both opening up and both having x-intercepts of negative 5 and 4the green parabola has a y-intercepts of negative 20 the red parabola has a y-intercepts of negative 40 and passes through the point (negative 4, negative 16).

Notice that $f (x) = x^{2} + x - 20$ and $f (x) = 2 x^{2} + 2 x - 40$ have the same zeros. These quadratic functions are related.

Only $f (x) = 2 x^{2} + 2 x - 40$ passes through $(- 4, - 16)$ and is the specific function we were looking for at the start.

It is very important to note that this is why, in the previous lesson, we were looking for “a function that has zeros at –5 and 4” and not “the function that has zeros at –5 and 4.”

This difference in wording is very important, since there are infinitely many functions that are related by their solutions.

We can only determine an exact specific function if we have the additional information of a point through which the function passes.

Let's try one more example.

Example 2

Write the quadratic function, $g (x)$ , in standard form, that has the zeros –7 and –3 and passes through the point $(- 2, - 5)$ .

Since the zeros are –7 and –3, the factored form of the function is $g (x) = (x + 7) (x + 3)$ .

The intercept form of the function is $g (x) = a (x + 7) (x + 3)$ .

We substitute values using the point $(- 2, - 5)$ to find the value of $a$ .

$g (x) = a (x + 7) (x + 3)$

$- 5 = a ((- 2) + 7) ((- 2) + 3)$

$- 5 = a (5) (1)$

$- 5 = 5 a$

$- 1 = a$

Substitute the value of $a$ into the intercept form of the function.

$g (x) = a (x + 7) (x + 3)$

$g (x) = - 1 (x + 7) (x + 3)$

$g (x) = - 1 (x^{2} + 7 x + 3 x + 21)$

$g (x) = - 1 (x^{2} + 10 x + 21)$

$g (x) = - x^{2} - 10 x - 21$

The specific function that has zeros –7 and –3 and passes through the point $(- 2, - 5)$ is:

$g (x) = - x^{2} - 10 x - 21$

Let's graph the functions to check our work.

Graph of two parabolas, one red and one green, both with x-intercepts of negative 7 and negative 3 the green parabola opens up the red parabola opens down and passes through the point (negative 2, negative 5).

Again, notice that $g (x) = x^{2} + 10 x + 21$ and $g (x) = - x^{2} - 10 x - 21$ have the same zeros. They are related quadratic functions.

Even though the leading coefficient is negative and the function has been reflected over the $x$ -axis, only $g (x) = - x^{2} - 10 x - 21$ passes through $(- 2, - 5)$ .

Try it

Try It: Finding a Quadratic Function from Its Zeros and a Point

Write the quadratic function, $h (x)$ , that has the zeros 2 and 6 and passes through the point $(7, 15)$ .

Here is how to find a function when given its zeros and a point:

The zeros are 2 and 6.

The intercept form is $h (x) = a (x - 2) (x - 6)$ .

Use the point $(7, 15)$ to find $a$ .

$h (x) = a (x - 2) (x - 6)$

$15 = a ((7) - 2) ((7) - 6)$

$15 = a (5) (1)$

$3 = a$

Substitute 3 for the value of $a$ in the intercept form.

$h (x) = a (x - 2) (x - 6)$

$h (x) = 3 (x^{2} - 2 x - 6 x + 12)$

$h (x) = 3 (x^{2} - 8 x + 12)$

$h (x) = 3 x^{2} - 24 x + 36$

The specific quadratic with zeros of 2 and 6 that passes through the point $(7, 15)$ is $h (x) = 3 x^{2} - 24 x + 36$ .

8.11.3 Finding a Quadratic Function from Its Zeros and a Point

Activity

Video: Finding a Quadratic Function from Its Zeros and a Point

Additional Resources

Finding a Quadratic Function from Its Zeros and a Point

Try It: Finding a Quadratic Function from Its Zeros and a Point