Find the vertex and axis of symmetry of the graph of the function.
-$$f ( x ) = x ^ { 2 } + 2 x - 3$$
A) (-1, -4); x = -1
B) (1, 4); x = 1
C) (1, -4); x = 1
D) (-1, 4); x = -1

Question

Quizplus · Accepted Answer

The vertex of a parabola in the form $f(x) = ax^2 + bx + c$ can be found using the formula $(- \frac{b}{2a}, f(- \frac{b}{2a}))$. For $f(x) = x^2 + 2x - 3$, $a = 1$ and $b = 2$, so the x-coordinate of the vertex is $- \frac{2}{2(1)} = -1$. Plugging this back into the function gives $f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$, making the vertex (-1, -4). The axis of symmetry is $x = - \frac{b}{2a} = -1$.

Find the Vertex and Axis of Symmetry of the Graph f(x)=x2+2x−3f ( x ) = x ^ { 2 } + 2 x - 3f(x)=x2+2x−3

Find the Vertex and Axis of Symmetry of the Graph $f ( x ) = x ^ { 2 } + 2 x - 3$