what is the vertex of the graph of the function f(x) = x^2 + 8x - 2?\n(-4, 18)\n(0, -2)\n(-8, -2)\n(-4, -18)

Explanation:

Step1: Identify coefficients

For the quadratic function $f(x)=ax^{2}+bx + c$, here $a = 1$, $b = 8$, $c=-2$.

Step2: Find x - coordinate of vertex

The formula for the x - coordinate of the vertex of a quadratic function is $x=-\frac{b}{2a}$. Substitute $a = 1$ and $b = 8$ into the formula: $x=-\frac{8}{2\times1}=-4$.

Step3: Find y - coordinate of vertex

Substitute $x = - 4$ into the function $f(x)=x^{2}+8x - 2$. Then $f(-4)=(-4)^{2}+8\times(-4)-2=16-32 - 2=-18$.

Answer:

D. (-4, -18)