Write an equation for a parabola with x-intercepts (-1,0) and (4,0) which passes through the point (3,-12)

Answer 1

To find the equation of a parabola, we can start by using the vertex form, which is given by:

y = a(x - h)^2 + k

where (h, k) is the vertex of the parabola.

We are given that the x-intercepts are (-1,0) and (4,0), which means the parabola crosses the x-axis at these points. Since the y-coordinate at the x-intercepts is 0, we can substitute these values into the equation:

0 = a(-1 - h)^2 + k
0 = a(4 - h)^2 + k

Next, we are given that the parabola passes through the point (3,-12). We can substitute these values into the equation as well:

-12 = a(3 - h)^2 + k

We now have a system of three equations:

1) 0 = a(-1 - h)^2 + k
2) 0 = a(4 - h)^2 + k
3) -12 = a(3 - h)^2 + k

To solve this system, we can use the elimination method. Subtracting equation 2) from equation 1), we get:

0 = a(-1 - h)^2 + k - a(4 - h)^2 - k
0 = a((-1)^2 + 2(-1)(h) + h^2) - a((4)^2 + 2(4)(h) + h^2)
0 = a + 2ah + ah^2 - a - 8a - 2ah - ah^2
0 = -6a - 2ah

Simplifying further, we have:

-6a - 2ah = 0
-2a(3 + h) = 0

Since a cannot be zero, we have:

h = -3

Substituting h = -3 back into equation 1), we can solve for k:

0 = a(-1 - (-3))^2 + k
0 = a(2)^2 + k
0 = 4a + k

k = -4a

Substituting h = -3 and k = -4a back into equation 3), we can solve for a:

-12 = a(3 - (-3))^2 + k
-12 = a(6)^2 + (-4a)
-12 = 36a - 4a
0 = 32a
a = 0

Since a = 0, the equation of the parabola is:

y = 0(x - (-3))^2 + (-4a)
y = -4a

Thus, the equation of the parabola is y = 0.