Question

Write the standard form of the equation of the parabola that has a vertex at (4, 3) and passes through the point (-4, -2).

Answer 1

My answer is y = -5/64(x - 4)^2 + 3

Answer 2

Actually, a1234, I'd be much more interested in seeing how you got your answer than I am in the answer itself.

The vertex form of the equation of a parabola is y=a(x-h)^2+k, where (h,k) is the vertex. This assumes that the parabola opens up or down, not to the right or to the left.

You are given the vertex and a certain point which lies on the parabola. Substituting these values into the equation above should enable you to calculate the value of "a".

Again, please share with the rest of us the steps you went through in trying to find "a."

Answer 3

y=a(x-h)^2+k
y=a(x-4)^2 + 3
And then to find "a," I put the other values into the equation.
-2 = a(-4 - 4)^2 + 3
-2 = a(-8)^2 + 3
-2 = 64a + 3
-5 = 64a
a = -5/64

Answer 4

@a1234 looks good, why did you post this? doesn't seem to be a question here :-)

Answer 5

Just getting the answer checked...thanks!

Answer 6

okay. you can check it yourself, of course. plug in the coordinates of the vertex and make sure the resulting statement is true:

\[y = -\frac{5}{64}(x-4)^2+3\]\[3=-\frac{5}{64}(4-4)^2+3\]\[3=0+3\]\[3=3\checkmark\]

and our known point (-4,-2):

\[-2=-\frac{5}{64}(-4-4)^2+3\]\[-2=-\frac{5}{64}(64)+3\]\[-2=-5+3\]\[-2=-2\checkmark\]

vertex is located symmetrically, so if vertex is at (4,3) and (-4,-2) is a point, then the mirror point must also work, so (8+4,-2) or (12,-2) should also be on the parabola:

\[-2= -\frac{5}{64}(12-4)^2+3\]\[-2=-\frac{5}{64}{64}+3\]\[-2=-2\checkmark\]
everything checks out!

here's a graph: