How do you sketch the graph of #y=(x-2)^2-7# and describe the transformation?

1 Answer
Sep 9, 2017

See below.

Explanation:

The function is in the form: #a(x - h )^2 +k#
Where #h# is the axis of symmetry, and #k# is the maximum or minimum value of the function. This is known as the vertex of the parabola.

From example: vertex is at #( 2 , -7 )#

We now need to find roots and #y# axis intercept. This will then give us a sufficient number of plotting points.

Expand #y = ( x - 2 )^2 - 7# , and equate it to #0#

#x^2 - 4x - 3 = 0#

Solution by quadratic formula gives roots:

#( 2 + sqrt7 , 0 )# and #( 2 - sqrt(7) , 0 )#

#y# axis intercept is where #x = 0#

#y = (0)^2 - 4(0) -3 #

# ( 0 , -3 ) #

So all plotting points are:

#( 2 , -7 )# ,#( 2 + sqrt7 , 0 )# , #( 2 - sqrt(7) , 0 )# , # ( 0 , -3 ) #

Graph:
graph{x^2 -4x -3 [-5, 10, -12.8, 20]}

This can be viewed as the graph of #y = x^2# translated 2 units to the right and 7 units in the #- y # direction.