# How do you find the axis of symmetry and vertex point of the function: y= -1.5x^2+6x?

Jun 22, 2018

Vertex: $\left(2 , 6\right)$

Axis of symmetry: $x = 2$

#### Explanation:

$y = - 1.5 {x}^{2} + 6 x$

To find the $x$-coordinate of the vertex of a standard quadratic equation (y = ax^2 + bx + c), we use the formula $\frac{- b}{2 a}$.

We know that $a = - 1.5$ and $b = 6$, so let's plug them into the formula:
$x = \frac{- 6}{2 \left(- 1.5\right)} = - \frac{6}{-} 3 = 2$

To find the $y$-coordinate of the vertex, just plug in the $x$-coordinate back into the original equation:
$y = - 1.5 {x}^{2} + 6 x$

$y = - 1.5 {\left(2\right)}^{2} + 6 \left(2\right)$

$y = - 1.5 \left(4\right) + 12$

$y = - 6 + 12$

$y = 6$

Therefore, the vertex is at $\left(2 , 6\right)$.

The axis of symmetry is the line of the $x$-coordinate of the vertex, so it's $x = 2$.

Here's a graph of this equation (desmos.com):

As you can see, the vertex is indeed at $\left(2 , 6\right)$.

Hope this helps!