Question

Find the equation of a parabola with the following conditions, vertex (0,-2), passing through (3,25)

Answer 1

Quiz!!!! Why does it say "a parabola", rather than "the parabola"?

Answer 2

The vertex form of parabola is: y = a(x - h)^2 + k ---- (1)
where (h, k) is the vertex and a is constant.
You are given (h, k). Plug that into (1).

The parabola passes through (3, 25).
Plug that into (1) and solve for a.

Answer 3

Or... Answer my question.

Answer 4

Not the original poster. It should be "the parabola" in the question since the point is given. If the point is not given, then it can be "a parabola" where we can pick any "a" value. It will be one of the many parabolas with the vertex at (0, -2).

Answer 5

That is what my teacher wrote on the paper, I didnt have anything to do with the wording.

Answer 6

Not so. There are at least two solutions. Let's assume, at this level, that the axis of symmetry is parallel to a coordinate axis.

\((y-k)^{2} = A(x-h)\)
\((y-k) = B(x-h)^{2}\)

Both valid solutions.

Impress your teacher by finding both solutions. I'm not asking you to take responsibility for wording. I'm asking you to solve the problem that is presented. You must first understand it.

Answer 7

True. It can be a vertical or a horizontal parabola.

Answer 8

i got 9a^2-23 or do I leave it at 25=9a^2-2 and she doesn't like people telling her that she is wrong or that there is more than one solution. She is very picky, but I'll try and see what she says.

Answer 9

Well, in this case, "very picky" = "simply wrong" if it is believed that there is a unique solution to the problem, as stated. Good luck with that. :-)

Answer 10

Thanks, I will need it for my test tomorrow. She counted a student wrong once because they didn't solve the problem the exact way she wanted but still got the same answer.. very picky but I try.

Answer 11

Assume it is a vertical parabola:
y = a(x - h)^2 + k
what do you get when you substitute for (h,k)?

Answer 12

y=a(x-0)^2-2 ?

Answer 13

Yes, which can be simplified to: y = ax^2 - 2
It passes through (3, 25)
put x = 3 and y = 25 into the equation and solve for a.

Answer 14

so 25=9a^2-2
then 9a

Answer 15

"a" is not squared in the equation.

Answer 16

9a^2=27
then 9a= (sq rt) 27
a= (sq rt 3)/3 ?

Answer 17

"a" is not squared in the equation.

Answer 18

y = ax^2 - 2
It passes through (3, 25)
Put x = 3 and y = 25 into the equation and solve for a.
25 = a(3)^2 - 2
25 = 9a - 2
25+2 = 9a
27 = 9a
a = 27/9 = 3

The equation of the parabola is: y = 3x^2 - 2.

Answer 19

\[y=a(x-h)^2+k\]where (h,k) is the vertex
1) Plug the vertex into the equation as well as the point given to solve for "a"
2) see @ranga 's algebraic steps
3) now with "a" known and the vertex, simplify the equation so it has the form y= x^2....

Answer 20

The standard equation is y = ax^2 + bx + c = 0.
There is min (or max) when x = -b/2a. The parabola 's vertex is at (0, -2), then b = 0.
The y-intercept, when x = 0, is the vertex itself, then c = -2.
Write that y passes through point (3, 25), to find a.
25 = 9a -2 --> 9a = 27 -> a = 3.
Finally: y = 3x^2 -2

Answer 21

Wandering about a bit, aren't we?

Two solutions with axis of symmetry parallel to a coordinate axis.

\((y+2)^{2} = Ax\)
\((25+2)^{2} = A(3) \implies 27^{2} = 3A \implies A = 3^{5} = 243\)

and

\((y+2) = Bx^{2}\)
\((25+2) = B(3)^{2} \implies 27 = 9B \implies B = 3\)