Expert: Paul Klarreich - 9/6/2007

Question
Paul, I have been stumped on this question for way too long and I dont know
where to start anymore. The problem comes out of a Calculus book, but its
reviewing old information from algebra. Heres the problem

Sketch the graph of the equation by translating, reflecting, compressing, and
sketching the graph of y= x^2 approximately.

y = -2(x+1)^2 -3

I just need to do the algebra, because graphing it is easy. Thank you in
advance.

Answer
Questioner:   Jason
Category:  Calculus
Private:  No
 
Subject:  Review For calculus
Question:  Paul, I have been stumped on this question for way too long and I dont know
where to start anymore. The problem comes out of a Calculus book, but its
reviewing old information from algebra. Heres the problem

Sketch the graph of the equation by translating, reflecting, compressing, and
sketching the graph of y= x^2 approximately.

y = -2(x+1)^2 -3

I just need to do the algebra, because graphing it is easy. Thank you in
advance.
.....................................
Hi, Jason,

I am not sure what you mean by "graphing it is easy."  I suppose you mean: "Plugging it into my TI84 is

easy."

But I think your teacher wants you to do this:

1. Sketch the graph of  y = x^2.  That's your basic parabola with its vertex at the origin.  NOW WORK FROM

THE INSIDE OUT to get the actual equation.

2. Replace  x  by  (x+1).  

New equation:  y = (x + 1)^2.
New graph:  Translated 1 unit to the left.  Vertex now at  (-1,0).

3. Replace  (x + 1)^2  by  2(x + 1)^2.  

New equation:  y = 2(x + 1)^2.
New graph:  Stretched vertically by a factor of 2.  Steeper and (seems to be) narrower.

4. Replace  2(x + 1)^2  by  -2(x + 1)^2.  

New equation:  y = -2(x + 1)^2.
New graph:  Inverted.  Vertex still at (-1,0) but this vertex is at the top of the graph, not the bottom.

5. Replace  -2(x + 1)^2  by  -2(x + 1)^2 - 3

New equation:  y = -2(x + 1)^2 - 3.
New graph:  Translated 3 units downward.  Vertex now at  (-1,-3), and this is the final graph.