# How do you find the vertex and axis of symmetry, and then graph the parabola given by:  f(x)=-4x^2 ?

Apr 7, 2018

The axis of symmetry is $x = 0$.

The vertex is $\left(9 , 0\right)$.

#### Explanation:

Given:

$f \left(x\right) = - 4 {x}^{2}$ is a quadratic equation in standard form:

$f \left(x\right) = a {x}^{2} + b x + c$,

where:

$a = - 4$, $b = 0$, $c = 0$.

The axis of symmetry and the $x$ coordinate of the vertex can be found using the formula:

$x = \frac{- b}{2 a}$

$x = \frac{0}{2 \cdot 0}$

$x = 0$

To find the $y$ coordinate of the vertex, substitute $y$ for $f \left(x\right)$ and $0$ for $x$.

$y = - 4 {\left(0\right)}^{2}$

$y = 0$

The vertex is $\left(0 , 0\right)$. Since $a < 0$, the vertex is the maximum point and the parabola opens downward.

There are no x- or y-intercepts. We can determine additional points by substituting values for $x$ and solving for $y$.

When $x = - \frac{1}{2}$,

$y = - 4 {\left(- \frac{1}{2}\right)}^{2}$

$y = - 4 \times \frac{1}{4}$

$y = - 1$

Point: $\left(- \frac{1}{2} , - 1\right)$

When $x = \frac{1}{2}$,

$y = - 4 {\left(\frac{1}{2}\right)}^{2}$

$y = - 4 \times \frac{1}{4}$

$y = - 1$

Point: $\left(\frac{1}{2} , - 1\right)$

When $x = - 1$,

$y = - 4 {\left(- 1\right)}^{2}$

$y = - 4 \times 1$

$y = - 4$

Point: $\left(- 1 , - 4\right)$

When $x = 1$,

$y = - 4 {\left(1\right)}^{2}$

$y = - 4 \times 1$

$y = - 4$

Point: $\left(1 , - 4\right)$

Plot the vertex and additional points and sketch a parabola through them. Do not connect the dots.

graph{y=-4x^2 [-10, 10, -5, 5]}