Question

what is the vertex of f(x) = 4x² + 8x - 2?
(8, -2)
(0, -6)
(-6, -1)
(-1, -6)

Explanation:

Step1: Identify coefficients

For the quadratic function $f(x)=ax^{2}+bx + c$, here $a = 4$, $b = 8$, $c=-27$.

Step2: Calculate x - coordinate of vertex

The formula for the x - coordinate of the vertex of a quadratic function is $x=-\frac{b}{2a}$. Substitute $a = 4$ and $b = 8$ into it: $x=-\frac{8}{2\times4}=-\frac{8}{8}=-1$.

Step3: Calculate y - coordinate of vertex

Substitute $x = - 1$ into the function $f(x)=4x^{2}+8x - 27$. Then $f(-1)=4\times(-1)^{2}+8\times(-1)-27=4 - 8 - 27=-6$.

Answer:

D. $(-1,-6)$