Answer:
Step-by-step explanation:
You have to solve this as a system. We have 3 coordinates given; the first 2 are given as zeros of the quadratic, the 3rd one is given as a regular coordinate. The zeros of a quadratic are where the y values of a coordinate are 0's because the x-intercepts of a quadratic (aka zeros) exist where y = 0. We use each of those points in the standard form of a quadratic to solve for a, b, and c. First, (0, 0):
[tex]0 = a(0)^2+b(0)+c[/tex] which gives us that c = 0. We'll use that value of c as we move forward with the problem. Next we'll use (3, 0):
[tex]0=a(3)^2+b(3)+0\\0 =9a+3b[/tex]Hold that thought for a minute or 2. Next we'll use the coordinate (1, 6):
[tex]6=a(1)^2+b(1)+0\\\\6=a+b[/tex] Now we have a system that we can solve using either substitution or elimination for a and b. I used elimination:
6 = a + b
0 = 9a + 3b and multiply the top equation by -3:
-18 = -3a - 3b
0 = 9a + 3b
so
-18 = 6a and
a = -3. Now back solve for b:
6 = -3 + b so
b = 9 and our quadratic is
[tex]y=-3x^2+9x+0[/tex] or just
[tex]y=-3x^2+9x[/tex]
