Question

write an equation for the parabola whose vertex is at (3,8) and which passes through (4,-3)

Answer 1

Plug (3,8) and (4,-3) into

\[y = a(x - h)^2 + k\]

Then solve for a

Answer 2

-3 = a(4 - 3)^2 + 8
-3 = a + 8
-8 - 3 = a
-11 = a

So the equation of the parabola is
\[y = -11(x - 3)^2 + 8\]

Answer 3

Thank you :)

Answer 4

vertex form is a(x-h)^(2)+k
(h,k) is the vertex, and a is what affects the parabola, (whether it widens, narrows, or flips)
so just plug in what you know. Drawing it could help too.


Since parabolas are symmetrical you would do the same thing on the other side, move over one, and down 11. (Sorry that my drawn is unproportional)
even being not to scale, it is obvious that the parabola is getting narrow, and flipped upside down, so absolute value of a>1 and negative.

plug in the vertex Y=a(x-3)^(2)+8
after this plug in the point you had been given (4,-3) into x and y spots

-3=a(4-3)^(2)+8
and solve for a. You already know it will be negative.

-3=a(1)+8
-8 -8

-11=a(1)
which would be -11=a

so then just plug everything into vertex form to get an ending result y=-11(x-3)^(2)+8.

This would be your final answer, unless asked to put it in standard form. But this is usually acceptable:)