Question

How do you find the vertex and the intercepts for `f(x)=2x^2-12x+21`?

Answer 1

`v=(6,21)` and there are no interceptions.

Explanation:

The vertex is the minimum of the function, so if we set the standard notation `f(x)=ax^2+bx+c` it is expressed as

`v_x=-b/{2a}=6` and `v_y=f(v_x)=21`

The intercepts are the roots of `f`, but `f` has no real roots since

`Delta/4=(b/2)^2-ac=36-42<0`

Answer 2

Y-intercept: `(0,21)`
Vertex: `(3,3)`
No x intercepts

Explanation:

`f(x)=2x^2-12x+21`

Y-intercept:
Find `f(0)`:
`f(0)=2(0)^2-12(0)+21=21`

Vertex- Complete the square of the equation
`f(x)=2[x^2-6x+10.5]`

`=2[(x-3)^2+1.5]`

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

Looking at the vertex form, the vertex is at `(3,3)`

X-intercepts:
Methods include using the quadratic formula, factoring, graphing, or completing the square/vertex form.
Using vertex form as found above:

`0=f(x)=2(x-3)^2+3`
`-3/2=(x-3)^2`
`x=+-sqrt(-3/2)+3`

This is a square root of a negative number, meaning the function has no `x` intercepts.