Question

How do you find the vertex and the intercepts for `f(x) = 4x^2 - 16x + 17`?

Answer 1

vertex: `(x,y)=(2,1)`
y-intercept: `17`
(no x-intercepts)

Explanation:

Rewriting the given equation: `f(x)=4x^2-16x+17`
as
`color(white)("XXX")f(x)=4(x^2-4x)+17`

`color(white)("XXX")f(x)=color(blue)(4)(x^2-4xcolor(red)(+4))+17color(blue)(-4)(color(red)(4))`

`color(white)("XXX")f(x)=4(x-2)^2+1`
which is the vertex form with vertex at `(2,1)`

The y-intercept is the value of `y` (i.e.`f(x)`) when `x=0`
`color(white)("XXX")f(0)=4(0-2)^2+1 =17`

The x-intercepts (if they exist) are the values of `x` for which `f(x)=0`
...but `(x-2)^2>=0` for all values of `x`
and therefore `4(x-2)^2+1 > 0` for all values of `x`.
`color(white)("XXX")f(x)!=0` for any value of `x`
graph{4x^2-16x+17 [-2, 9.096, -0.598, 4.95]}