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

1 Answer
Jun 25, 2018

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.