# How to graph a parabola y=3/2x^2 - 3x-5/2?

Jun 21, 2015

Complete the square to find the vertex, axis of symmetry, etc.

#### Explanation:

$\frac{3}{2} {\left(x - 1\right)}^{2} = \frac{3}{2} \left({x}^{2} - 2 x + 1\right) = \frac{3}{2} {x}^{2} - 3 x + \frac{3}{2}$

So

$y = \frac{3}{2} {x}^{2} - 3 x - \frac{5}{2}$

$= \frac{3}{2} {\left(x - 1\right)}^{2} - \frac{3}{2} - \frac{5}{2}$

$= \frac{3}{2} {\left(x - 1\right)}^{2} - 4$

This is in vertex form,

allowing us to tell that the vertex is at $\left(1 , - 4\right)$

and the axis of symmetry is $x = 1$.

The intersection with the $y$ axis is at $\left(0 , \frac{3}{2}\right)$

The intersections with the $x$ axis are where $y = 0$, so

$0 = \frac{3}{2} {\left(x - 1\right)}^{2} - 4$

Add $4$ to both sides to get:

$\frac{3}{2} {\left(x - 1\right)}^{2} = 4$

Multiply both sides by $\frac{2}{3}$ to get:

${\left(x - 1\right)}^{2} = \frac{2}{3} \cdot 4$

Hence $x = 1 \pm 2 \sqrt{\frac{2}{3}} \cong 1 \pm 1.633$

So the intersections with the $y$ axis are approximately:

$\left(- 0.633 , 0\right)$ and $\left(2.633 , 0\right)$

graph{3/2x^2-3x-5/2 [-10, 10, -5, 5]}