Question

find an equation of the form y-k = a(x-h)^2 for the parabola with vertex (-1,5) and y intercept 8

Answer 1

Recall: \[y=a(x-h)^{2}+k\] is vertex form for a parabola, where (h,k) is the vertex.

So if you need the vertex to be (-1,5), you can plug those values in for h and k.
\[y=a(x-h)^{2}+k\]
\[y=a(x-(-1))^{2}+(5)\]
and simplify
\[y=a(x+1)^{2}+5\]

From here, we're going to have to manipulate the a value in order to get this thing to have a y-intercept of 8.

As it sits right now, the y-intercept is made by taking:
\[a(h)^{2}+k\]
If you want to design it so that it goes through the y axis at 8, you have to set it equal to 8 and solve for a.
\[a(-1)^{2}+5=8\]
simplify
a+5=8
subtract 5 on both sides
a = 3

Now, we have one final manipulation to get it back to the form they asked for it in.

\[y=3(x+1)^{2}+5\]
You want to put the 5 on the same side as the y, so you can subtract it off of both sides.

So, finally:
\[y-5=3(x+1)^{2}\]

I hope that helps!! :)

Answer 2

yesss thankss

Answer 3

If the idea of \[a(h)^{2}+k \]
is a little too weird, you can always force the a value out by starting with the equation as we have it at that point:
\[y=a(x+1)^{2}+5\]
And plugging in the point that we need to be the vertical intercept (0,8) for x and y.
\[(8)=a((0)+1)^{2}+5\]
If you solve for a here, you'll get the same result. This is a little more traditional too. :)

So first, simplify
\[8=a(1)^{2}+5\]

\[8=1a+5\]

\[8=a+5\]
Then subtract 5 from both sides, to get:
3=a