# How do you graph the quadratic function and identify the vertex and axis of symmetry for y=y=-3x^2+5?

Jul 3, 2017

Vertex $= \left(0 , + 5\right)$
Axis of symmetry $x = 0$

#### Explanation:

For a quadratic in standard form: $a {x}^{2} + b x + c$
The axis of symmetry and vertex will occur where $x = \frac{- b}{2 a}$

In our example: $- 3 {x}^{2} + 5 \to a = - 3 , b = 0 , c = + 5$

$\frac{- b}{2 a} = \frac{0}{-} 6 = 0$

Hence the axis of symmetry is $x = 0$ [i.e. the $y -$axis]

And the y-coordinate of the vertex of the parabola is:
$0 + 5 = 5$

$\therefore$ Vertex $= \left(0 , + 5\right)$

Also, since the coefficient of ${x}^{2} < 0 \to y \left(0\right) = 5$ is the absolute maximum of $y$.

The vertex and axis of symmetry can be seen on the graph of $y$ below.

graph{-x^2+5 [-11.25, 11.25, -5.63, 5.62]}