Transforming quadratic functions

[Nelo]

Homework Statement

Sketch the graph of y=x^2 and graph y=1/2(x+4)^2 -5

Homework Equations

The Attempt at a Solution

So, This is the method I use for vertical compression/expansion

Step pattern of a parabola at y=x^2 is 1,3,5

Start at vertex, go out by one, go up by 1 and plot the point. From that point, you go out by 1 go up by 3, plot the point. then go out by one, go up by 5 and plot the point.. use symmetry to get the other side and that gives you y=x^2 of a porabola.

If you have something like y=4(x-2) +4

then the vertex is (2,4) , and since there is a vertical stretch you multiply the step pattern of, 1,3,5 by 4 , giving you, 4 ,12, 20, ... effectively... Start at the vertex of 2,2 . and go out by 1 and up by 4, plot the point, then go out by one up by 12, plot the point. and so forth, that is what I've learned as the step pattern for a parabola. when you have vertica lstretch you multiply by 4.

Why doesn't this work now? Usually when you have a graph like 1/2(x-4)^2 +5
You go to the left 4 units, up 5 units, then multiply the step patern of 1,3,5 by 0.5. But for some reason that doesn't work. So how the hell do you do a vertical stretch on the parabola?

 

[rock.freak667]

Why doesn't this work now? Usually when you have a graph like 1/2(x-4)^2 +5
You go to the left 4 units, up 5 units, then multiply the step patern of 1,3,5 by 0.5. But for some reason that doesn't work. So how the hell do you do a vertical stretch on the parabola?

for the (x-4)² part, you'd need to shift the graph of x² four (4) units to the right. Then you'd need to stretch (x-4)² by the factor of ½ and then the vertical translation.

I've never really learned transformation graphs as 1,3,5 so I am not sure if I am telling you something you already know.

 

[eumyang]

Why doesn't this work now? Usually when you have a graph like 1/2(x-4)^2 +5

Earlier, you wrote y = 1/2(x + 4)² - 5. Which one is it? I'm going to assume that you mean
y = 1/2(x - 4)² + 5

You go to the left 4 units, up 5 units, then multiply the step patern of 1,3,5 by 0.5. But for some reason that doesn't work. So how the hell do you do a vertical stretch on the parabola?

It does work. But you have to go the right 4 units and up 5 units to plot the vertex. So if you plot the points
(4, __)
(5, __)
(6, __)
(7, __)
(I'll let you fill in the blanks), then you'll see that the step pattern (multiplied by 1/2) is applied.

Don't forget that regarding transformations, if we start with y = f(x), then
y = f(x - c) is a transformation to the right by c units, and
y = f(x + c) is a transformation to the left by c units.
I've seen a lot of students get those mixed up.