Question

How do you graph `y=x^2+8-3x`?

Answer 1

Here's one way: complete the square to find its vertex, then calculate a few more points by plugging in values of `x`.

Explanation:

Let's rearrange the equation:
`y=x^2-3x+8`

This equation can't be factored, so let's complete the square:
`y = (x^2 - 3x + 9/4 - 9/4) + 8`
`y = (x - 3/2)^2 - 9/4 + 8`
`y = (x-3/2)^2+23/4`

This is the equation in vertex form: `y=a(x-h)^2+k`
We know the vertex is `(h,k) = (3/2, 23/4).`

The leading coefficient `a` is positive, which means that the parabola opens upwards.

We can get a few more points of the parabola by plugging in some values of `x` around `3/2`.

Substituting `x=2`, we find `(2,6)`.
Substituting `x=1`, we find `(1,6)`.
Substituting `x=5/2`, we find `(5/2, 27/4)`.
Substituting `x=1/2`, we find `(1/2, 27/4)`.

Graph and connect these points. Be sure to label the equation of the graph, label the axes, and include arrowheads.