Question 1206303: Which quadratic function in vertex form can be represented by the graph that has the vertex at (-6, 0) and passes through the point (2, 8)?
A. y = -1/8x^2 - 6
B. y = 1/8(x + 6)^2
C. y = 1/8x^2 - 6
D. y = -1/8(x + 6)^2
Found 3 solutions by greenestamps, Solver92311, MathTherapy:
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
Vertex form is  or 
All the answer choices are in the second form. So with vertex (-6,0), the correct answer choice has to be in the form  , or  .
Answer choices B and D are both in that form. In answer choice B, a is 1/8; in answer choice D, a is -1/8. Since the vertex is at (-6,0) and the graph passes through (2,8), we know that the parabola opens upward, which means a is positive, so answer choice B is correct.
More formally, we can substitute x=2 in answer choices B and D to see which gives us the correct value of 8 for y. With answer choice B, x=2 gives us y=8, which is correct; with answer choice D, x=2 gives us y=-8, which is not. So again answer choice B is correct.
ANSWER: B
Answer by Solver92311(821)  (Show Source):
Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website!
Which quadratic function in vertex form can be represented by the graph that has the vertex at (-6, 0) and passes through the point (2, 8)?
A. y = -1/8x^2 - 6
B. y = 1/8(x + 6)^2
C. y = 1/8x^2 - 6
D. y = -1/8(x + 6)^2
y = a(x - h)2 + k <==== Vertex form of a quadratic function
8 = a(2 - - 6)2 + 0 --- Substituting (- 6, 0) for (h, k), and (2, 8) for (x, y)
8 = a(8)2
8 = 64a
y = a(x - h)2 + k
Substituting  for a, and (- 6, 0) for (h, k) gives us: 
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