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Question 737392: A parabola has vertex at (4, 6) and intersect the x-axis at x = 8. Find its equation in the form of y = ax^2 + bx + c
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! A parabola has vertex at (4, 6) and intersect the x-axis at x = 8. Find its equation in the form of y = ax^2 + bx + c
.
Vertex form of a parabola:
y = a(x – h)^2 + k
where
(h,k) is the vertex
.
Plugging in the vertex at (4,6)
y = a(x – 4)^2 + 6
.
Since it intersects the x-axis at x=8, this point is at (8,0).
Plugging it in the the equation, we can solve for 'a':
0 = a(8 – 4)^2 + 6
0 = a(4)^2 + 6
0 = a(16) + 6
-6 = a(16)
-6/16 = a
-3/8 = a
.
Plug above back into:
y = a(x – 4)^2 + 6
y = (-3/8)(x – 4)^2 + 6
.
Now, we rearrange terms to get the desired form of:
y = ax^2 + bx + c
.
y = (-3/8)(x – 4)^2 + 6
y = (-3/8)(x – 4)(x – 4) + 6
y = (-3/8)(x^2-8x+16) + 6
y = (-3/8)x^2 - (-3/8)8x + (-3/8)16 + 6
y = (-3/8)x^2 - (-3)x + (-3)2 + 6
y = (-3/8)x^2 + 3x - 6 + 6
y = (-3/8)x^2 + 3x (this is what they're looking for)
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