Question

How do you find the vertex and intercepts for `y = x^2 - 4x`?

Answer 1

See below.

Explanation:

Quick answer:

`x`-intercepts:
Solve `x^2-4x = 0`

`x(x-4) = 0`,

`x = 0,4`

`y`-intercept:
When `x = 0`, the equation gives us `y = 0`

Vertex: is midway between the intercepts. The midpoint is the average.

`x = (0+4)/2 = 2`

When `x = 2`, the equation gives us `y = (2)^2-4(2) = 4-8=-4`

The vertex is `(2,-4)`

Answer 2

A different method that uses `color(red)("part of")` completing the square. Purely given to show that sometimes different methods work.

Explanation:

Given: `y=x^2-4x`

`color(blue)("Determine y-intercept")`

Write as `y=x^2-4x+0`
`" "color(red)(uarr)`
`" "color(red)("y-intercept")`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine vertex")`

Note that in the standardised form of `y=ax^2+bx+c` our equation is such that `a=1` so `b/a=(-4)/1=-4`

`x_("vertex")=(-1/2)xxb/a->(-1/2)xx(-4)=+2`

By substitution

`y_("vertex")=(2)^2-4(2)" "=" "4-8" "=" "-8`

Vertex`->(x,y)=(2,-8)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine x-intercept")`

The `x^2` term is positive so the graph is of form `uu` thus has a minimum. `y_("vertex")<0` so x-intercepts exist.

Factorising

The common term is `x` so lets factor that out giving.

`y=x(x-4)`

The x-intercepts are at `y=0` so by substitution

`y=0=x^2(x-4)`

Thus `x=0 and 4`