Expert: Jack Cheng - 11/15/2006

Question
How do I draw a rough sketch of this quadratic function?  Also, I need to label the vertex and the x and y intercepts.

f(x)=3x^2-12x

Answer
Hi Susan,

To graph this function, let's first get a faint idea as to what it looks like first.

1) The coefficient of the x^2 term is positive, so the parabola opens up.

2) Let's now find the intercepts. The y-intercept is the value of the function when x=0. Plugging 0 into the function, it's also 0 (which means it's also one of the x-intercepts).

The x-intercept is the value of x when y=0 (or f(x)=0, in this case). So:

0 = 3x^2-12x
0 = 3x(x-4)
so x = 0 or 4

3) Now that we know that the x-intercepts are 0 and 4, we know that the x-coordinate of the vertex must be 2 (it's the middle of the x-intercepts; this only works when you have two x-intercepts). Plugging that in, we find the y-coordinate to be 3*2^2-12*2 = 12-24 = -12. So, the vertex is (2, -12)

4) Now that we know the vertex and two points of the graph, we basically almost have the graph. You can plug in some more values of x to make the graph more apparent.

I hope this helps,
Jack

P.S. You can also do what is called Completing the Square to find an equivalent equation that is easier to graph (i.e. it tells you the vertex immediately). Look more into it if you are interested.