Question

How do you find the intercepts, vertex and graph `f(x)=-2x^2+8x-3`?

Answer 1

Vertex `(2, 5)`
X intercept `(3.58, 0); (0.42, 0)`
Y-Intercept `(0, -3)`

Explanation:

Given -

`f(x) = -2x^2+8x-3`

Vertex

`x=(-b)/(2a)=(-8)/(2 xx -2)=8/(-4)=-2`

At `x=2`

`y=-2(2^2)+8(2)-3`
`y=-8+16-3=16-11=5`

Vertex `(2, 5)`

X intercept

`-2x^2+8x-3=0`

{Completing the square]

`x^2-4x+3/2=0` [divide all the terms by `-2`]

`x^2-4x=-3/2` [Take the constant term to the right]
`x^2-4x+4=-3/2+4` (divide the coefficient of `x` square it and add to both sides]
`x^2-4x+4=-3/2+4=(-3+8)/2=5/2`
`(x-2)^2=-5/2`
`x-2=+-sqrt(5/2)=+-1.58`

`x=1.58+2=3.58`
`x=-1.58+2=0.42`

X intercept `(3.58, 0); (0.42, 0)`

Y-Intercept
At `x=0`

`y=-2(0)^2+8(0)-3`
`y=-3`

Y-Intercept `(0, -3)`

To graph the function take a few values on either side of `x=2`
Find the corresponding `y` values.
Tabulate them. Graph the pairs