Question 1176690: Find the quadratic function y=f(x) whose graph has a vertex (-2,3) and passes through the point (-5,0).Write the function in standard form.
Answer by Solver92311(821)   (Show Source):
You can put this solution on YOUR website!
Since y is a function of x, we are dealing with a graph that is a parabola with a vertical axis of symmetry. Since the vertex is the point ) , the equation of the axis of symmetry is  . The axis of symmetry intersects the  -axis at the point ) . The distance from the other given point, ) , to the axis is |\ =\ 3) . Hence, because of symmetry, another point on the graph must exist on the -axis 3 units from the intersection of the axis of symmetry and the -axis, to wit, the point )
The standard form of a quadratic function is  .
Since the -value of  in the desired function must yield the  -value  , we can say:
^2\ +\ b(-5)\ +\ c\ =\ 0)
And, using the data from the other two points:
^2\ +\ b(-2)\ +\ c\ =\ 3)
and
^2\ +\ b(1)\ +\ c\ =\ 0)
Simplifying:
Solve the 3X3 linear system for the a, b, c, coefficients of your desired function.
Alternate solution
Start with the vertex form of a parabola:
^2\ +\ k)
Insert the coordinates of the vertex for )
\)^2\ +\ 3)
^2\ +\ 3)
Now, insert the coordinates of the other given point, , and solve for 
^2\ +\ 3\ =\ 0)
Back to our vertex form:
^2\ +\ 3)
All that is left is expanding the squared binomial and collecting like terms to produce the desired standard form.
John
My calculator said it, I believe it, that settles it.
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