Question

eoc style question
a quadratic function, $f(x)$, has the $x$-intercepts $(-7,0)$ and $(6,0)$. the point $(-5,-110)$ lies on the parabola.
which of the following is the equation of $f(x)$ in standard form?
$circ$ $f(x)=5x^2 - 210$
$circ$ $f(x)=x^2 + x - 42$
$circ$ $f(x)=5x^2 + 5x - 210$
$circ$ $f(x)=\frac{55}{6}x^2 - \frac{55}{6}x - 385$

Explanation:

Step1: Write factored form template

$f(x)=a(x+7)(x-6)$

Step2: Substitute point (-5,-110)

$-110=a(-5+7)(-5-6)$

Step3: Calculate right-hand side

$-110=a(2)(-11) \implies -110=-22a$

Step4: Solve for $a$

$a=\frac{-110}{-22}=5$

Step5: Expand to standard form

$f(x)=5(x+7)(x-6)=5(x^2+x-42)=5x^2+5x-210$

Answer:

$\boldsymbol{f(x)=5x^2+5x-210}$ (corresponding to the third option)