Question

How do you graph `F(x)=x^2-6x+8`?

Answer 1

See answer below

Explanation:

Given: `f(x) = x^2 -6x + 8`

This function is a quadratic - a graph of a parabola.

Factor and let `f(x) = 0` to find the x-intercepts :

`x^2 -6x + 8 = (x - 2)(x - 4) = 0`

#x - 2 = 0 " "=> x = 2; " "x - 4 = 0 " "=> x = 4

`x`-intercepts: " "(2, 0), (4, 0)#

Find the y-intercept let `x = 0`:

`f(0) = 0^2 -6*0 + 8 = 0`

`y`-intercept: " "(0, 8)#

Find the vertex . When the equation is in `Ax^2 + Bx + C = 0`,

the vertex is `(-B/(2A), f(-B/(2A)))`

`-B/(2A) = 6/2 = 3`

`f(3) = 3^2 -6*3 + 8 = -1`

vertex: `(3, -1)`

Plot a couple of other points using point-plotting. Since `x` is the independent variable, you can select any `x` and calculate the corresponding `y`:

`ul(" "x" "|" "y" ")`
`" "1" "|" "3" "`
`" "5" "|" "3" "`
`" "6" "|" "8" "`

graph{x^2 -6x + 8 [-5, 10, -2, 10]}