SOLUTION: A building has an entry the shape of a parabolic arch 22 ft high and 28 ft wide at the base as shown below. Find an equation for the parabola if the vertex is put at the origin

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A building has an entry the shape of a parabolic arch 22 ft high and 28 ft wide at the base as shown below. Find an equation for the parabola if the vertex is put at the origin      Log On


   

Question 982003: A building has an entry the shape of a parabolic arch 22 ft high and 28 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.
Please Help!
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Vertex expected at (0,0).
y=a%28x-0%29%5E2%2B0 using standard form.

a%3C0 because concave downward.

Sketching the graph helps.
y value at origin is -22, and because 28 ft wide, 14 feet from line of symmetry to either end at the base. Two points on the parabola are (14,-22) and (-14,-22). You can find "a" using either of these points.
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Sketch the graph and think about that discussion.


FIND COEFFICIENT a.
y=ax%5E2
a=y%2Fx%5E2
a=%28-22%29%2F%2814%29%5E2
a=-22%2F196
cross%28a=-11%2F108%29, a=-11%2F98

FINISHED FINDING EQUATION
cross%28y=-%2811%2F108%29x%5E2%29, highlight%28y=-%2811%2F98%29x%5E2%29


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graph(300,300,-16,16,2,-30,-11x^2/98)
Rendering tags are not included because the graph was shown incorrectly, no matter how the function was adjusted. Use a graphing tool to see the graph properly. Go to google search engine and just put y=-(11/98)*x^2 into the search text field and click ENTER or GO. You should see the graph correctly.