# How do you find the vertex and axis of symmetry, and then graph the parabola given by: y = -2x^2 + 10x - 1?

Oct 16, 2015

Vetex $\left(\frac{5}{2} , \frac{- 23}{2}\right)$

Axis of symmetry - $x = \frac{5}{2}$

#### Explanation:

$- 2 {x}^{2} + 10 x - 1$

Find the vertex

$x = \frac{- b}{2 a} = \frac{- 10}{2 \times \left(- 2\right)} = \frac{- 10}{- 4} = \frac{5}{2}$

$y = - 2 {\left(\frac{5}{2}\right)}^{2} + 10 \left(\frac{5}{2}\right) - 1 = - 2 \left(\frac{25}{4}\right) + 25 - 1$
$y = - \left(\frac{25}{2}\right) + 25 - 1 = \frac{- 25 + 50 - 2}{2} = \frac{- 23}{2}$

Vetex $\left(\frac{5}{2} , \frac{- 23}{2}\right)$

Axis of symmetry -

$x = \frac{5}{2}$

Since co-efficent of ${x}^{2}$ is negative, the curve opens downwards. The function has a maximum at the vertex.

Take a few points less than $x = 2.5$ and a few points greater.
Calculate their corresponding $y$ values.

x y
0 -1
1 7
2 11
2.5 11.5
3 11
4 7
5 -1
Plot the points and join them with a smooth curve.

graph{-2x^2+10x-1 [-32.48, 32.48, -16.24, 16.22]}