SOLUTION: A Parabola Passes Through The Point -1,5 And Its Vertex Is At 0,-4 With A Horizontal Axis Of Symmetry. What Is The Equation Of The Parabola?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A Parabola Passes Through The Point -1,5 And Its Vertex Is At 0,-4 With A Horizontal Axis Of Symmetry. What Is The Equation Of The Parabola?      Log On


   

Question 1203843: A Parabola Passes Through The Point -1,5 And Its Vertex Is At 0,-4 With A Horizontal Axis Of Symmetry. What Is The Equation Of The Parabola?
Found 2 solutions by MathLover1, josgarithmetic:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
given:
Passes through the point (-1,5)
Vertex is at (0,-4) => h=0, k=-4
If a parabola has a horizontal axis, the standard form of the equation of the parabola is this:
%28y+-+k%29%5E2+=+4p%28x+-+h%29
substitute h=0, k=-4
%28y+-+%28-4%29%29%5E2+=+4p%28x+-+0%29
%28y+%2B4%29%5E2+=+4p%2Ax

use given point to calculate 4p
%285+%2B4%29%5E2+=+4p%2A%28-1%29
81+=+-4p
4p=-81

equation is:
%28y+%2B4%29%5E2+=+-81x





Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
y-k=a%28x-h%29%5E2 if vertical symmetry axis
x-h=a%28y-k%29%5E2 if horizontal symmetry axis; vertex at (h,k).

Given vertex (0,-4)
x=a%28y%2B4%29%5E2

Given included point on graph (-1,5)
a=x%2F%28y%2B4%29%5E2
a=-1%2F%285%2B4%29%5E2
a=-1%2F81


Equation for the parabola
highlight%28x=-%281%2F81%29%28y%2B4%29%5E2%29




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(Not carefully rechecked)