SOLUTION: Could you please help me with the equation of a parabola? Please see the attached picture of the graph. Thank you in advance! <a rel=nofollow HREF="https://ibb.co/YN8RCjT"><im

Algebra ->  Graphs -> SOLUTION: Could you please help me with the equation of a parabola? Please see the attached picture of the graph. Thank you in advance! <a rel=nofollow HREF="https://ibb.co/YN8RCjT"><im      Log On


   

Question 1192447: Could you please help me with the equation of a parabola? Please see the attached picture of the graph. Thank you in advance!
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Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


One form of the equation for a parabola with vertex (h,k) that opens upward is

y=%281%2F%284p%29%29%28x-h%29%5E2%2Bk

With the equation in that form, p is the directed distance (i.e., might be negative) from the vertex to the focus.

Your graph shows the vertex at (0,2); and it shows that the directed distance from the vertex to the focus is +1. So p=1 and the equation is

y=%281%2F%284%29%29%28x%29%5E2%2B2

You can use the graph to verify the equation by using x=2 or x=-2 and seeing that for both of those x values the value of y is 3.

Outside the scope of your question, note that 4p=4 is the length of the latus rectum -- the segment that is perpendicular to the axis of the parabola through the focus, with endpoints on the parabola.