Question

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

Answer 1

Vetex `(5/2, (-23)/2)`

Axis of symmetry - `x=5/2`

Explanation:

`-2x^2+10x-1`

Find the vertex

`x=(-b)/(2a)=(-10)/(2 xx(-2))=(-10)/(-4)=5/2`

`y=-2(5/2)^2+10(5/2)-1=-2(25/4)+25-1`
`y=-(25/2)+25-1 = (-25+50-2)/2=(-23)/2`

Vetex `(5/2, (-23)/2)`

Axis of symmetry -

`x=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]}