Question

the axis of symmetry for a function in the form $f(x)=x^2 + 4x - 5$ is $x = -2$. what are the coordinates of the vertex of the graph? $(-2, -17)$ $(-17, -2)$ $(-9, -2)$ $(-2, -9)$

Explanation:

Step1: Recall vertex x - coordinate

The axis of symmetry of a parabola \(y = ax^{2}+bx + c\) is \(x=-\frac{b}{2a}\), and the x - coordinate of the vertex is the same as the axis of symmetry. Here, the axis of symmetry is \(x = - 2\), so the x - coordinate of the vertex is \(x=-2\).

Step2: Find y - coordinate of vertex

Substitute \(x=-2\) into the function \(f(x)=x^{2}+4x - 5\).
\[

$$\begin{align*} f(-2)&=(-2)^{2}+4\times(-2)-5\\ &=4-8 - 5\\ &=4-(8 + 5)\\ &=4 - 13\\ &=-9 \end{align*}$$

\]
So the coordinates of the vertex are \((-2,-9)\).

Answer:

\((-2, - 9)\)