SOLUTION: A parabola y = ax^2 + bx + c has vertex (4, 2). If (2, 0) is on the parabola, then find the value of abc

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Question 258319: A parabola y = ax^2 + bx + c has vertex (4, 2). If (2, 0) is on the parabola, then find the value of abc
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The formula for the x position of the vertex is
x=-b%2F2a
-b%2F2a=4
b=-8a
Now using the points,
y+=+ax%5E2+%2B+bx+%2B+c+
2+=+16a+%2B+4b+%2B+c+
16a%2B4b%2Bc=2
.
.
y+=+ax%5E2+%2B+bx+%2B+c+
0+=+4a+%2B+2b+%2B+c+
4a%2B2b%2Bc=0
.
.
.
Now substitute for b.
b=-8a
16a%2B4b%2Bc=2
16a-32a%2Bc=2
1.-16a%2Bc=2
.
.
.
4a%2B2b%2Bc=0
4a-16a%2Bc=0
2.-12a%2Bc=0%7D%7D%0D%0A%7B%7B%7Bc=12a
Substitute into eq. 1 to solve for a.
-16a%2Bc=2
-16a%2B12a=2
-4a=2
a=-1%2F2
Then,
c=12a
c=-6
and finally,
b=-8a
b=4
So then,
y+=+%28-1%2F2%29x%5E2+%2B+4x+-6+

abc=%28-1%2F2%29%284%29%28-6%29=12