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Question

i need help with this question....

Find the equation of the parabola that satisfies the conditions.

The Parabola has zeroes at -2 and 6. It passes through the point (5,3)

Answer 1

If the parabola has zeroes at -2 and 6, then you have points at (-2, 0) and (6, 0). You also have a third point (5, 3) that we can use.

The general form of a parabola is (assuming that this opens up or down):

y = ax^2 + bx + c

We can plug in points and find the constants a, b, and c:

(-2, 0): 0 = a(-2)^2 + b(-2) + c
0 = 4a - 2b + c

(6, 0): 0 = a(6)^2 + b(6) + c
0 = 36a + 6b + c

(5, 3): 3 = a(5)^2 + b(5) + c
3 = 25a + 5b + c

We can solve this using some fancy matrix operations, but we'll do this the old fasioned way - eliminating variables and substituting. If we rewrite the first equation in terms of c, we have c = 2b - 4a. We can start with that and plug in to the second and third equations:

2: 0 = 36a + 6b + c
0 = 36a + 6b + (2b - 4a)
0 = 32a + 8b

3: 3 = 25a + 5b + c
3 = 25a + 5b + (2b - 4a)
3 = 21a + 7b

We can rewrite the new second equation in terms of b:
0 = 32a + 8b
8b = -32a
b = -4a

And now we can substitute this into the new third equation:
3 = 21a + 7b
3 = 21a + 7(-4a)
3 = 21a - 28a
3 = -7a
a = -3/7

From here, we can just go back and plug in and find b and c:

b = -4a
b = -4(-3/7)
b = 12/7

c = 2b - 4a
c = 2(12/7) - 4(-3/7)
c = 24/7 + 12/7
c = 36/7

You equation ends up being:

y = (-3/7)x^2 + (12/7)x + 36/7