SOLUTION: The graph of a quadratic function passes through the origin and has vertex (4, 16). Use symmetry to identify one other point that must be on the graph of this function. Fin

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The graph of a quadratic function passes through the origin and has vertex (4, 16). Use symmetry to identify one other point that must be on the graph of this function. Fin      Log On


   

Question 1179834: The graph of a quadratic function passes through the origin and has
vertex (4, 16).
Use symmetry to identify one other point that must be on the graph of
this function.
Find an equation for this function.
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Find an equation for this function.
if a function passes through the origin, we have a point (0,0)
if the vertex (4, 16), we know that h=4 and k=16
using vertex form, we have
y=a%28x-h%29%5E2%2Bk..........use given points to calculate a
0=a%280-4%29%5E2%2B16
0=16a%2B16
-16=16a
a=-1
your function is:
y=-%28x-4%29%5E2%2B16
Use symmetry to identify one other point that must be on the graph of
this function.
symmetry axis passes through vertex, so use line of symmetry
x=4
P(-2,-20) and Q(10,-20) are both 6 units from the line of symmetry





Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
The graph of a quadratic function passes through the origin and has
vertex (4, 16).
Use symmetry to identify one other point that must be on the graph of
this function.
Find an equation for this function.
~~~~~~~~~~~~~ 


The other point is the point symmetrical to the origin point (0,0) relative to the axis of symmetry x=4 of the parabola.


This symmetrical point is 4 units to the right relative the symmetry line x=4,

and it has the same y-coordinate of 0.


So, the other point is (8,0).


The equation of the parabola in the vertex form is


    y = a%2A%28x-4%29 + 16.     (1)


Since it is going through the origin point  (x,y) = (0,0),  from equation (1) we have


    0 = a%2A%280-4%29%5E2 + 16,   or

    0 = 16a + 16

    16a = -16,

      a = -16%2F16 = -1.


Thus the equation of the parabola is  

      y = -%28x-4%29%5E2+%2B+16.



    


    Plot y = -%28x-4%29%5E2%2B16.


Comparing my solution with that by @MathLover1,  notice  the difference:

            the problem asks to find the point on the plot,  symmetric to the origin of coordinates.

            The origin of coordinates is the point  (0,0),  and the point symmetrical to it is the point  (8,0),  as  I  responded.

            @MathLover1 considers totally different points,  that are  IRRELEVANT  to the posed question.