Identify all the points that are on the following quadratic function: y = x2 - 3x (0,3) (0,0) (2,-2) (-1,4) (-3,15)

If we want to see if a particular point is on the quadratic function we need to replace each pair [tex](x,y)[/tex] in the quadratic function expression and check the equality.

For example :

We have the following point : [tex](x,y)=(1,2)[/tex]

If we replace by [tex]x=1[/tex] and [tex]y=2[/tex] in the quadratic function :

[tex]y=x^{2}-3x[/tex]

[tex]2=1^{2}-3(1)[/tex]

[tex]2=1-3[/tex]

[tex]2=-2[/tex]

The final expression is false, therefore the point [tex](1,2)[/tex] is not on the quadratic function.

Now let's work with the points we were given.

The first point is [tex](0,3)[/tex] ⇒ If we replace in the quadratic function :

[tex]y=x^{2}-3x[/tex]

[tex]3=0^{2}-3(0)[/tex]

[tex]3=0[/tex]

We conclude that the point [tex](0,3)[/tex] is not in the quadratic function.

The second point is [tex](0,0)[/tex]

[tex]y=x^{2}-3x[/tex]

[tex]0=0^{2}-3(0)[/tex]

[tex]0=0[/tex]

This point is on the quadratic function.

The third point is [tex](2,-2)[/tex]

If we replace in the quadratic function :

[tex]y=x^{2}-3x[/tex]

[tex]-2=(2)^{2}-3(2)[/tex]

[tex]-2=4-6[/tex]

[tex]-2=-2[/tex]

The point [tex](2,-2)[/tex] is on the quadratic function.

The fourth point is [tex](-1,4)[/tex]

[tex]y=x^{2}-3x[/tex]

[tex]4=(-1)^{2}-3(-1)[/tex]

[tex]4=1+3[/tex]

[tex]4=4[/tex]

Therefore the point [tex](-1,4)[/tex] is on the quadratic function.

Finally, we have the point [tex](-3,15)[/tex]

If we replace in the quadratic function :

[tex]y=x^{2}-3x[/tex]

[tex]15=(-3)^{2}-3(-3)[/tex]

[tex]15=9+9[/tex]

[tex]15=18[/tex]

This point is not on the quadratic function.

We conclude that the points [tex](0,0),(2,-2)[/tex] and [tex](-1,4)[/tex] are on the quadratic function [tex]y=x^{2}-3x[/tex]