# What is the axis of symmetry and vertex for the graph f(x)=-4x^2 ?

Jun 2, 2018

See below

#### Explanation:

The axis of symmetry can be calculated for a quadratic in standard form ($a {x}^{2} + b x + c$) by the equation $x = - \frac{b}{2 a}$
In the equation in your question, $a = - 4 , b = 0$, and $c = 0$. Thus, the axis of symmetry is at $x = 0$:
$x = - \frac{b}{2 a} = - \frac{0}{2 \cdot - 4} = \frac{0}{-} 8 = 0$

To find the vertex, substitute the x-coordinate of the axis of symmetry for x in the original equation to find its y-coordinate:
$y = - 4 {x}^{2} = - 4 \cdot {0}^{2} = - 4 \cdot 0 = 0$

So the axis of symmetry is $x = 0$ and the vertex is at $\left(0 , 0\right)$.