#f(x)=x^2#. The graph of #f# is shifted vertically #3# up, and then scaled vertically by a factor of #5#. The result is the graph of #ax^2+bx+c# where #a = 5, b= 0# and #c=# ?

1 Answer
David B.
Sep 15, 2017

#c=3# assuming that the vertical scaling factor applied is referenced to the vertex of the parabola and not the origin.

Explanation:

The simplest way to solve this question is to think of the #y#-intercept of the graph, in other words the point where #x=0#. We start with the graph of #y=x^2# which is then shifted up by #3#. This implies that our new #y#-intercept is at #y=3#. In coordinate form, this is the point #(0,3)# which is also the vertex of the parabola.

This assumes that the vertical scaling that is then applied is done so to the parabola while leaving the vertex at #(0,3)#. Therefore, we can simply solve the equation

#y = ax^2+bx+c#

where

#y=3, x=0, a=5# and #b=0#

#=> c = 3#

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