Question

How do you graph the parabola `y= -3x^2 + 12x - 8` using vertex, intercepts and additional points?

Answer 1

See details below

Explanation:

Every polinomic expresion of second degree is a parabola with general expresion `y=ax^2+bx+c`

The interception point with `y` axis we can obtain it appliying the value `x=0`. We obtain `y=-8`. So, our parabola inercepts
to `y` axis at `(0,-8)`

The intercetions with `x` axis we can obtain them appliying the value `y=0`. We get

`0=-3x^2+12x-8` we will obtain two points wich are the roots of that polinomical equation. In our case

`x=(-b+-sqrt(b^2-4ac))/(2a)=(-12+-sqrt(144+96))/-6`

`x=2-2/3sqrt15` and `x=2+2/3sqrt15`

So the intercets points with `x` axis are `(2-2/3sqrt15,0)` and `(2+2/3sqrt15,0)`

The vertex is given by `x=-b/(2a)=-12/-6=2` in wich value `y` reaches the value `y=4`

See the graph and zoom it (with mouse wheel) and pass the mouse cursor by intercept points, You will see the values we already get

graph{-3x^2+12x-8 [-22.99, 22.64, -16.14, 6.66]}