## The problem

Click to see the original problem

I need help with This,

"Find a quadratic equation with x-intercepts -3 and 2 and a y-intercept of 12. Put it in standard form" I don't know how to use your site to get the answer.

thank you

## Answer provided by our tutors

The standard form of quadratic equation is:
'y = ax^2+bx+c',
where 'a', 'b' and 'c' are constants and 'x' and 'y' are variables.

From the text of the problem we know that the line goes trough (-3, 0),(2, 0) and (0, 12).​
This means that we can plug the values for 'x' and 'y' to find 'a', 'b' and 'c'.​

For (x, y) = (-3, 0) we have:​
a(-3)^2 + b*(-3) + c = 0​
9a - 3b + c = 0

For (x, y) = (2, 0) we have:​
a(2)^2 + b*2 + c = 0​​
4a + 2b + c = 0

For (x, y) = (0, 12) we have:​​
a*0^2 + b*0 + c = 12
c = 12

If we plug c = 12 into ​'9a - 3b + c = 0​' and '4a + 2b + c = 0​' we get a system of equations:​

9a - 3b + 12 = 0​​
4a + 2b + 12 = 0​​
......
Click here to see the system of equations solved for 'a' and 'b'

a = -2

b = -2

The standard form of quadratic equation is:

y = -2x^2 - 2x + 12

If we graph the equation we see that the line indeed goes trough the -3, 0),(2, 0) and (0, 12): 