# How do you find the axis of symmetry and vertex point of the function:  y=2x^2-3x+2?

Oct 10, 2015

Axis of symmetry is $x = \frac{3}{4}$, vertex is $\left(\frac{3}{4} , \frac{7}{8}\right)$

#### Explanation:

Convert x terms into a perfect square as follows:

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

$y = 2 \left({x}^{2} - \frac{3 x}{2} + \frac{9}{16} - \frac{9}{16}\right) + 2$

$y = 2 \left({x}^{2} - \frac{3 x}{2} + \frac{9}{16}\right) - \frac{9}{8} + 2$

$y = 2 {\left(x - \frac{3}{4}\right)}^{2} + \frac{7}{8}$. This equation is now in the vertex form. Axis of symmetry is $x = \frac{3}{4}$, vertex is $\left(\frac{3}{4} , \frac{7}{8}\right)$