We need a few things to graph this: the xx and yy intercepts and the "slope". Because xx is squared, I know that this will be a quadratic function. There aren't slopes for quadratics, but we can look for certain points.
First, let's look for yy-intercepts:
y=ax^2+bx+color(red)(c)y=ax2+bx+cIn our equation (y=3x^2)(y=3x2), we don't have a last constant, so our yy-intercept is 00.
Now let's look for our xx-intercept. To find it, we set y=0y=0 and solve for xx:
0=3x^20=3x2
0=x^20=x2
sqrt(0)=sqrt(x^2)√0=√x2
x=0x=0
So, our xx and yy intercepts are both 00, which means our vertex is (0,0)(0,0)
Now we have two out of our three required pieces. Now let's think this next one through...
If we start at (0,0)(0,0) and move up one, our x=1x=1:
y=3(1)^2y=3(1)2
y=3y=3
That means our point is (1, 3)(1,3).
Now let's solve for when x=-1x=−1:
y=3(-1)^2y=3(−1)2
y=3y=3
So, our second point is (-1,3)(−1,3)
We can solve for more points this way, but for the most part, having three reference points to draw from are enough.
Our vertex is (0,0)(0,0), and our next two points (which will help dictate the "slope") are (-1,3)(−1,3) and (1,3)(1,3)
graph{y=3x^2}
