Question

How do you find the vertex of `y= x^2-8x+5`?

Answer 1

The simplest method for finding the vertex of the given polynomial is to convert it into "vertex form" to find the vertex at `(4,-11)`

Explanation:

Given `y=x^2-8x+5`

If we can convert this into vertex form:
`color(white)("XXXX")` `y=m(x-a)^2+b`
`color(white)("XXXX")` `color(white)("XXXX")`for some constants `a` and `b`
then the vertex will be at `(a,b)`

The easiest way to do this conversion is via a "completion of the square method"

If `x^2-8x` are the first 2 terms of a squared binomial
`color(white)("XXXX")` `color(white)("XXXX")`note: `(m+n)^2 = (m^2+2mn+n^2)`
then the third term needs to be `((-8x)/(2x))^2 =16`

So we add (and then subtract `16` to "complete the square"

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

Simplifying:
`y =(x-4)^2 + (-11)`
`color(white)("XXXX")`which is our desired "vertex form" (with `m=1`)

So the vertex is at `(4,-11)`

Answer 2

Find vertex of f(x) = x^2 - 8x + 5

Ans: Vertex (4, -11)

Explanation:

x-coordinate of vertex: `x = -b/(2a) = 8/2 = 4`

y-coordinate of vertex: `y = f(4) = 4^2 - 8(4) + 5 = - 11`