Question

How do you graph and identify the vertex and axis of symmetry for `5/2x(x-3)`?

Answer 1

If `y=5/2x(x-3)` then you can put it into the form of `y=ax^2+bx+c` to create a table of points, then plot them on the graph.

But since you've already factorised it, you can find the solutions (x-intercepts)
`0=5/2x(x-3)`
`0=5/2x` or `0=x-3`
`x=0` or `x=3`

The x-intercepts of your graph along with your vertex can allow you to plot a substantially accurate graph.

To find the vertex, complete the square:
`y=5/2x(x-3)`
`y=5/2x^2-15/2x`
`y=5/2(x^2-3x)`
`y=5/2(x^2-3x+(3/2)^2-(3/2)^2)`
`y=5/2((x-3/2)^2-9/4)`
`y=5/2(x-3/2)^2-45/8`

Your vertex then becomes `(3/2, -45/8)`
which is the same as `(1.5, -5.625)`
And your axis of symmetry `x=1.5`