Answer:
[tex]y=3x^2-12x-135[/tex]
Step-by-step explanation:
The standard form of a quadratic is [tex]y=ax^2+bx+c[/tex]
We will use the x and y values from each of our 3 points to find a, b, and c. Filling in the x and y values from each point:
First point (-5, 0):
[tex]0=a(-5)^2+b(-5)+c[/tex] and
0 = 25a - 5b + c
Second point (9, 0):
[tex]0=a(9)^2+b(9)+c[/tex] and
0 = 81a + 9b + c
Third point (8, -39):
[tex]-39=a(8)^2+b(8)+c[/tex] and
-39 = 64a + 8b + c
Use the elimination method of solving systems on the first 2 equations to eliminate the c. Multiply the first equation by -1 to get:
-25a + 5b - c = 0
81a + 9b + c = 0
When the c's cancel out you're left with
56a + 14b = 0
Now use the second and third equations and elimination to get rid of the c's. Multiply the second equation by -1 to get:
-81a - 9b - c = 0
64a + 8b + c = -39
When the c's cancel out you're left with
-17a - 1b = -39
Between those 2 bolded equations, eliminate the b's. Do this by multiplying the second of the 2 by 14 to get:
56a + 14b = 0
-238a - 14b = -546
When the b's cancel out you're left with
-182a = -546 and
a = 3
Use this value of a to back substitute to find b:
56a + 14b = 0 so 56(3) + 14b = 0 gives you
168 + 14b = 0 and 14b = -168 so
b = -12
Now back sub in a and b to find c:
0 = 25a - 5b + c gives you
0 = 75+ 60 + c so
0 = 135 + c and
c = -135
Put that all together into the standard form equation to get
[tex]y=3x^2-12x-135[/tex]
