Question

if $y = ax^2 + bx + c$, find $a, b, c$ if the zeros are $-2$ and $4$ and the $y$-intercept is $-4$.

Explanation:

Step1: Use roots to form equation

If the zeros are $x=-2$ and $x=4$, the quadratic can be written as $y=a(x+2)(x-4)$

Step2: Expand the factored form

$y=a(x^2 - 4x + 2x - 8)=a(x^2 - 2x - 8)$

Step3: Use y-intercept to find $a$

The y-intercept is when $x=0$, $y=-4$. Substitute:
$-4=a(0^2 - 2(0) - 8)$
$-4=-8a$
$a=\frac{-4}{-8}=\frac{1}{2}$

Step4: Substitute $a$ to find $b,c$

Substitute $a=\frac{1}{2}$ into $y=a(x^2 - 2x - 8)$:
$y=\frac{1}{2}x^2 - x - 4$
Compare to $y=ax^2+bx+c$ to get $b=-1$, $c=-4$.

Answer:

$a=\frac{1}{2}$, $b=-1$, $c=-4$