A building has an entry the shape of a parabolic arch 96 ft high and 18 ft wide at the base as shown below.
A parabola opening down with vertex at the origin is graphed on the coordinate plane. The height of the parabola from top to bottom is ninety six feet and its width from left to right is eighteen feet.
Find an equation for the parabola if the vertex is put at the origin of the coordinate system.

- anonymous

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- mathstudent55

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- anonymous

ok

- anonymous

whats do i do now?

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## More answers

- mathstudent55

\(y = ax^2 + bx + c\)
Use each of the three points in the equation above to come up with three equations.
Then solve the equations simultaneously for a, b, and c.

- anonymous

how do i use that formula u gave me?

- anonymous

because its at the origin, and its pointing down, the y would be the height

- mathstudent55

\(0 = a\cdot 0^2 + b \cdot 0 + c\)
c = 0

- anonymous

so y= -96, a=0?

- anonymous

o ok

- mathstudent55

That was for point (0, 0). We have c = 0.

- anonymous

ok

- anonymous

So y = 96,
since it's axis of symmetry is the origin, the x's on either side would be -9 and 9
SO we have now two points on the parabola. (9,96) and (-9,96)

- anonymous

(96 = a*-9^2 + b*-9 + 0)
(96 = a*9^2 + b*9 + 0)
these would be the two equations?

- mathstudent55

Now let's try the point (9, -96)
\(y = ax^2 + bx\) There is no c bec we already know c is zero.
\(-96 = a(9^2) + b(9) \)
\(81a + 9b = -96\)
Here is one equation.

- anonymous

2 equations? there will be only one equation

- mathstudent55

Now we use the point (-9, -96) to get the other equation.
\(-96 = a(-9)^2 + b(-9) \)
\(-96 = 81a - 9b\)
\(81a - 9b = -96\)
Here is the other equation. Since now we only have the two variables a and b, all we need is 2 equations. We can now find a and b.
We need to solve the system of equations below:
\(81a + 9b = -96\)
\(81a - 9b = -96\)

- anonymous

The vertex form, as you probably know, is:
y = a(x - h)^2 + k
where (h, k) is the vertex and that is point (0, 0), so your equation becomes:
y = a(x - 0)^2 + 0
or
y = ax^2
to find "a", you have to realize that (9, -96) is a point, so:
-96 = a(9^2) = 81a
a = -96/81 = -32/27
so,
y = (-32/27)x^2
vertex form:
y = (-32/27)(x - 0)^2 + 0

- anonymous

im confused now?

- anonymous

from there you can just expand the vertex form to get the standard form

- mathstudent55

Add the equations to eliminate b:
\(162a = -192\)
\(a = -\dfrac{192}{162} = -\dfrac{32}{27} \)
Now we need b.

- anonymous

I have no idea what math student is doing, we do not need2 equations!!!

- anonymous

ok can i get run by this step by step because im so lost

- anonymous

using vertex form first then expanding to get standard form is quite simple. Follow my previous long post

- anonymous

The vertex form, as you probably know, is:
y = a(x - h)^2 + k
where (h, k) is the vertex and that is point (0, 0), so your equation becomes:
y = a(x - 0)^2 + 0
or
y = ax^2
to find "a", you have to realize that (9, -96) is a point, so:
-96 = a(9^2) = 81a
a = -96/81 = -32/27
so,
y = (-32/27)x^2
vertex form:
y = (-32/27)(x - 0)^2 + 0
Then multiply it out to get the standard form.

- anonymous

OH i now see what math student is doing. Albeit a bit lengthy, but works

- mathstudent55

\(81a + 9b = -96\)
\(81\left( -\dfrac{32}{27} \right) + 9b = -96\)
\(-2592 + 243b = -2592\)
\(243b = 0\)
\(b = 0\)
Now we can write the equation of the parabola as:
\(y = ax^2 + bx + c\)
\(y = -\dfrac{32}{27}x^2 \)

- anonymous

so from y = (-32/27)(x - 0)^2 + 0 the standard form would be y = (-32/27)x^2?

- anonymous

Essentially the same thing, but using vertex method then expanding takes fewer time (just a hint for next time)

- anonymous

Yes chris

- anonymous

so that would be my equation for the parabola if the vertex is put at the origin of the coordinate system?

- anonymous

yes, that is what we both got.

- anonymous

perfect thank you very much

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