SOLUTION: Can you please explain to me how you create an equation (function) from looking at a parabola? I can make a parabola from an equation. But not the other way around. I need very eas

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Can you please explain to me how you create an equation (function) from looking at a parabola? I can make a parabola from an equation. But not the other way around. I need very eas      Log On


   

Question 41253: Can you please explain to me how you create an equation (function) from looking at a parabola? I can make a parabola from an equation. But not the other way around. I need very easy and explained steps!
Thanks for your time and effort.
-C.
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
Look at the parabola below.
graph%28+200%2C300%2C-1%2C7%2C-1%2C19%2C2%2B%28x-3%29%5E2%2C2+%29

At first find the vertex of the parabola from the figure.
You can easily do this.
Here the coordinates of the vertex are (3,2).
Also you can clearly see that the parabola is having an axis parallel to the x-axis.

If the coordinates of the vertex of a parabola with axis parallel to x-axis be (h,k) then its equation is of the form
y-k+=+4%2Aa%2A%28x-h%29%5E2__________________(1)

In this case h = 3, k = 2
Put this values in equation (1) then you get
y-2+=+4%2Aa%2A%28x-3%29%5E2__________________(2)

Now, note the coordinates of the point where the parabola cuts one of the coordinate axes.
Here the parabola cuts the y-axis at (0,11).
So (0,11) must be a point on the parabola and hence must satisfy its equation.
Hence x=0 and y=11 satisfies equation (2).
Therefore, 11-2+=+4%2Aa%2A%280-3%29%5E2
or 9+=+4%2Aa%2A3%5E2+=+36a
or a+=+1%2F4
Put a+=+1%2F4 in equation and get the required equation of the given parabola.
Here, the equation of the given parabola is y-2+=+%28x-3%29%5E2