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)
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)
Answer
Explanation:
Step1: Recall vertex - x formula
For a quadratic function $y = ax^{2}+bx + c$, the x - coordinate of the vertex is given by $x=-\frac{b}{2a}$. In the function $f(x)=x^{2}+8x - 2$, $a = 1$, $b = 8$, and $c=-2$. Then $x=-\frac{8}{2\times1}=-4$.
Step2: Find y - coordinate of vertex
Substitute $x = - 4$ into the function $f(x)=x^{2}+8x - 2$. So $f(-4)=(-4)^{2}+8\times(-4)-2=16-32 - 2=-18$.
Answer:
D. (-4, -18)