Question

How do find the vertex and axis of symmetry, and intercepts for a quadratic equation `y = -3x^2 + 12x - 8`?

Answer 1

Convert the equation into vertex form. Use the vertex and parabolic form to determine the axis of symmetry. Set one variable to `0` to determine the intercept of the other.

Explanation:

Conversion to Vertex Form
`color(white)("XXXX")` vertex form is `y=m(x-a)^2+b` with vertex at `(a,b)`
`y=-3x^2+12x-8`
`rarr` `color(white)("XXXX")` `y = (-3)(x^2-4x) -8`
`rarr` `color(white)("XXXX")` `y =(-3)(x^2-4x+4) -8 +12`
`rarr` `color(white)("XXXX")` `y = (-3)(x-2)^2+4`
The vertex is at `(2,4)`

Axis of Symmetry
The equation is a parabola in standard position (vertical axis) opening downward. So the axis of symmetry is a vertical line running through the vertex:
`color(white)("XXXX")` `x=2`.

Intercepts
`color(white)("XXXX")` y-intercept
Set `x=0` in equation to determine the y-intercept:
`y = 3(0)^2+12(0)-8`
`y=-8`
`color(white)("XXXX")` the y-intercept is `(-8)`

`color(white)("XXXX")` x-intercept
Set `y=0` and solve for `x` to determine the x-intercepts
`0 = 3x^2+12x-8`
Use the quadratic formula `x = (-b+-sqrt(b^2-4ac))/(2a)`
to determine x-intercepts

`x " intercepts" = (-12+-sqrt(144 +96))/6`

`color(white)("XXXX")` `= -2+- sqrt(240)/6`

`color(white)("XXXX")` `= -2+- (2sqrt(15))/3`