Question

A building has an entry the shape of a parabolic arch 96 ft high and 18 ft wide at the base, as shown below.
Find an equation for the parabola if the vertex is put at the origin of the coordinate system.

Answer 1

@jdoe0001

Answer 2

alrite
so.... well'll use the vertex form of a parabola, that is \(\bf y=a(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}\\
x=a(y-{\color{blue}{ k}})^2+{\color{brown}{ h}}\qquad\qquad vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\)

notice one thing
the parabola is opening downwards, and is vertical
downwards means, our leading term coefficient will be negative
and vertical, means, that the "x" variable is the squared one
thus, we'll end up using \(\large y=a(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}\)

Answer 3

Here:
y = ax^2 + bx + c

Use (0, 0):
0 = a(0^2) + b(0) + c
c = 0

Use (-9, -96)
-96 = a(-9)^2 + b(-9)
81a -9b = -96

Use (9, -96)
-96 = a(9)^2 + b(9)
81a +9b = -96


Solve the system of equations by addition:
81a +9b = -96
81a -9b = -96

162a = -192
a = -192/162 = -32/27

81(-32/27) + 9b = -96

this is as far as I got

Answer 4

hmmm

Answer 5

well... I was going to say... let's just take a peek at the graph :)

Answer 6

yeah but I have to explain for the assignment

Answer 7

this I guess would be the next step
81(-32/27) + 9b = -96
9b = 0
b = 0

y=?

Answer 8

so... lemme do the graph bit, which seems you already know the points to be used anyway



so... let us just pick say 9,-96

when x = 9, y = -96
we know that much from the graph, and we also know the vertex is at 0,0
so let us plug that in then in our parabola form
\(\bf y=a(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}
\\ \quad \\
-96=a(9-{\color{brown}{ 0}})^2+{\color{blue}{ 0}}\)

so... the only missing term is "a",
but solving that for "a", what does that give us?

well, surely you can find "a", and then just plug it in back in the vertex form equation :)

Answer 9

anyhow.. check your work :)
you already found "a"
so, just plug it in I gather :)