Question

what are the coordinates of the vertex of the function $f(x)=x^2 + 10x - 3$?
$\circ$ $(-5, -28)$
$\circ$ $(-5, 28)$
$\circ$ $(5, -28)$
$\circ$ $(5, 28)$

Explanation:

Step1: Find x-coordinate of vertex

For $f(x)=ax^2+bx+c$, $x=-\frac{b}{2a}$. Here $a=1, b=10$, so $x=-\frac{10}{2(1)}=-5$.

Step2: Find y-coordinate of vertex

Substitute $x=-5$ into $f(x)$: $f(-5)=(-5)^2+10(-5)-3=25-50-3=-28$.

Step3: Identify vertex coordinates

Combine x and y values: $(-5, -28)$

Answer:

A. (-5, -28)