Question

Can someone helpp pls

Answer 1

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):

$0 = a(0)^2+b(0)+c$ 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):

$0=a(3)^2+b(3)+0\\0 =9a+3b$Hold that thought for a minute or 2. Next we'll use the coordinate (1, 6):

$6=a(1)^2+b(1)+0\\\\6=a+b$ 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

$y=-3x^2+9x+0$ or just

$y=-3x^2+9x$