Question

How do you find the vertex and intercepts for `y =1/2x^2 +2x - 8`?

Answer 1

Vertex: `(-2,-10)`
y-intercept: `-8`
x-intercepts `-2(1+-sqrt(5))`

Explanation:

Finding the vertex
Notice that our target will be the vertex form `y=color(green)m(x-color(red)a)^2+color(blue)b` with vertex at `(color(red)a,color(blue)b)`

Given
`y=1/2x^2+2x-8`

Isolate the terms involving `x`
`y+8=1/2x^2+2x`

Modify so the coefficient of `x^2` is `1`
`2(y+8)=x^2+4x`

Noting that if `x^2+4x` are the first two terms of an expanded binomial `(x+a)^2=(x^2+2ax+a^2)` then `4x=2ax rArr a^2=2^2`
So we will need to add `2^2` (to both sides) to "complete the square"
`2(y+8)+2^2=x^2+4x+2^2`
or
`2(y+8)+4=(x+2)^2`

Now we reverse the process to isolate `y` and leave the result in vertex form:
`2(y-8)=(x+2)^2-4`

`(y-8)=color(green)(1/2)(x+2)^2-2`

`y=color(green)(1/2)(x+2)^2-10`

`y=color(green)(1/2)(x-color(red)(""(-2)))^2+color(blue)(""(-10))`

which is the vertex form with vertex at `(color(red)(-2),color(blue)(""(-10)))`

y-intercept
The y-intercept is the value of `y` when `x=0`

`y=1/2x^2+2x-8` with `x=0`
`color(white)("XXX")rArr y_((x=0)) =-8`

x-intercepts
Similarly the x-intercept values are the values of `x` for which `y=0`

So we need to solve `1/2x^2+2x-8=0`
or (simplified by multiplying both sides by `2`
`color(white)("XXX")x^2+4x-16=0`

Using the quadratic formula (ask if you need this), we can find
`color(white)("XXX")x=-2(1+-sqrt(5))`

graph{(1/2)x^2+2x-8 [-13.55, 11.76, -10.93, 1.73]}