Question

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

Explanation:

Step1: Recall vertex formula for parabola

For a quadratic function \( f(x) = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). In the function \( f(x)=x^{2}+8x - 2 \), we have \( a = 1 \), \( b = 8 \) and \( c=-2 \).

Step2: Calculate x - coordinate of vertex

Substitute \( a = 1 \) and \( b = 8 \) into the formula \( x=-\frac{b}{2a} \). We get \( x=-\frac{8}{2\times1}=- 4 \).

Step3: Calculate y - coordinate of vertex

Substitute \( x=-4 \) into the function \( f(x)=x^{2}+8x - 2 \). So \( f(-4)=(-4)^{2}+8\times(-4)-2=16-32 - 2=16-(32 + 2)=16 - 34=-18 \).

Answer:

(-4, -18)