The rainbow bridge of Utah is a natural arch that is approximately parabolic in shape. The arch is about 88m high. It is 84m across the base. Determine a quadratic relation, in standard form, that models the shape of the arch

Question

My answer was y=-22/441x^2+88
But the textbook said the answer is y=-22/441x^2+88/21x

I was wondering why is that the answer.

Answer 1

Depends where you place the arch on your graph.
I centered mine so the top is (0,88) and its x-intercepts are (42, 0) and (-42,0)

so using the vertex form of the parabola equation, I get
y = ax^2 + 88
but (42,0) lies on it
0 = 1764a + 88
a = -88/1764 = -22/441

which yields the same answer as yours.

If we place one end at (0,0) and the other at (84,0)
the vertex is (42,88) and we get
y = a(x-42)^2 + 88
sub in (0,0)
0 = 1764a + 88
the same a = -22/441
so y = -22/441(x - 42)^2 + 88
which when expanded yields their answer
(I was hoping for that)

Poorly worded question, since they should have indicated where the vertex was to be placed.

Answer 2

Would my answer be wrong if I used it on a test?

Answer 3

no, it would be correct if you included a sketch.
I would give you full marks.

Answer 4

The x intercepts or zeros are split equally between the y-axis, so x = -42, and x = 42 We can assume the centre of the bridge is at the origin (0, 0) with its sides contacting the ground (x-axis) equally split on either side of the "y-axis". So just 84/2, or half of "84". These points would form your "zero's", which will give you the necessary info. to get started. The top of the arch (or vertex) would be "88" meters up on the y-axis. hint: The Axis of sym would simply be at x = 0 zeros: -42, 42 vertex: (0, 88) y = a(x – 42)(x + 42) solve for "a" by subbing in the point (0, 88) for x and y last step, sub in your "a" and zeros then FOIL the brackets and simplify to standard form i.e. y = ax^2 + bx + c

Answer 5

Okay, here is the thing, the answer indeed what is written in the textbook
The answer is (-22/441x^2) + (88/21x)

We obviously have to use the quadratic form of a(x-r)(x-s) in which r and s are the zeros of the parabola. For my parabola i used (0,0) and (84,0).
This makes the equation for to be y= a(x)(x-84):
all i have to do now is solve for a, i used the vertex of the parabola which is 42, 88:
After insert x and y into the equation, the value of a comes to be -88/1764.
So the new quation is -88/1764(x)(x-84)
Now simplify to get:
-22/441(x^2+84x)
and simplify further to get (-22/441x^2) + (88/21x)
in reality the answer is that...and the question was worded fine hello!!!, u were supposed to use 0,0 and 84,0, u probably used the wrong stuff...