8.11.3 Finding a Quadratic Function from Its Zeros and a Point

Finding a Quadratic Function from Its Zeros and a Point

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

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

• The solutions are $x=1$ and $x=2$.

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

• Factor out the GCF, 2.

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

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 $a$.

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

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

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

$-16=-8a$

$2=a$

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

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

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

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

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

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

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

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

Let's graph the functions to check our work.

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

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

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 $a$.

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

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

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

$-5=5a$

$-1=a$

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

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

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

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

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

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

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

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

Let's graph the functions to check our work.

Again, notice that $g(x)=x^2+10x+21$ and $g(x)=-x^2-10x-21$ 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 $g(x)=-x^2-10x-21$ passes through $(-2,-5)$.