Question

Write a quadratic function in standard form that passes through (−6,0), (4,0)
, and (−4,16)
.(1 point)
f(x)=

Answer 1

To write a quadratic function in standard form \( f(x) = ax^2 + bx + c \) that passes through the points \((-6, 0)\), \((4, 0)\), and \((-4, 16)\), we start by using the roots given by the points \((-6, 0)\) and \((4, 0)\).

The roots of the quadratic function mean we can express the function in factored form:
\[
f(x) = a(x + 6)(x - 4)
\]
Here, \(a\) is a constant that we will determine using the third point \((-4, 16)\).

Now, we can expand this:
\[
f(x) = a((x + 6)(x - 4)) = a(x^2 + 2x - 24)
\]
So,
\[
f(x) = ax^2 + 2ax - 24a
\]

Next, we plug in the third point \((-4, 16)\) into the equation:
\[
16 = a(-4)^2 + 2a(-4) - 24a
\]
This simplifies to:
\[
16 = 16a - 8a - 24a
\]
\[
16 = 16a - 32a
\]
\[
16 = -16a
\]
Now we solve for \(a\):
\[
a = -1
\]

Now, we can substitute \(a\) back into the quadratic function:
\[
f(x) = -1(x^2 + 2x - 24) = -x^2 - 2x + 24
\]

Thus, the quadratic function in standard form is:
\[
\boxed{f(x) = -x^2 - 2x + 24}
\]