The graph of is shown below. Assume the domain of is and that the vertical spacing of grid lines is the same as the horizontal spacing of grid lines. **Part (a):** The points and are on the graph of Find and **Part (b):** Find the graph of Verify that your points from part (a) are on the graph. **Part (c):** The points and are on the graph of Find and **Part (d):** Find the graph of Be sure to verify that your points from part (c) are on the graph both algebraically and geometrically.

Please include graphs in your answer! Thank you!

Guest Nov 16, 2014

#1**+5 **

**If it makes you feel any better I struggled with this one - but I think it is right. **

**Part a y=f(2a)**

f(-4)=4 but f(2a)=-4 so 2a=-4 so a=-2

f(4)=-4 but f(2b)=4 so 2b=4 so b=2

**Part c y=f(2a-8)**

f(-4)=4 but F(2c-8)=-4 so 2c-8=-4 so 2c=4 so c=2

f(4)=-4 but f(2d-8)=4 so 2d-8=4 so 2d=12 so d=6 but this is outside the domain so it does not exist

Melody
Nov 17, 2014

#1**+5 **

Best Answer

**If it makes you feel any better I struggled with this one - but I think it is right. **

**Part a y=f(2a)**

f(-4)=4 but f(2a)=-4 so 2a=-4 so a=-2

f(4)=-4 but f(2b)=4 so 2b=4 so b=2

**Part c y=f(2a-8)**

f(-4)=4 but F(2c-8)=-4 so 2c-8=-4 so 2c=4 so c=2

f(4)=-4 but f(2d-8)=4 so 2d-8=4 so 2d=12 so d=6 but this is outside the domain so it does not exist

Melody
Nov 17, 2014

#2**0 **

Hi all,

Supermanaccz has asked me to come back an talk about this ancient question.

Thanks for your interest Supermanaccz

Firstly the question is not displaying properly because the coding used on this site has changed a lot in the past 3 and a bit years.

So I will try to put the question up how it may originally have been. I believe a diagram like the one below would have been included in the original question. Thie is the graph of y=f(x).

I have drawn it in a program called GeoGebra. GeoGebra is a great tool. I found it a little hard to use at first (especially for more complicated things) but if you persist with it it will become much easier. Geogebra is a free download. I use it often.

**Now to the actual question:**

The graph of is shown below. Assume the domain of is [-4,4] and that the vertical spacing of grid lines is the same as the horizontal spacing of grid lines.

Part (a): The points (a, 4) and (b, -4) are on the graph of y=f(2x) Find a and b

From the graph you can see that 2 points on the f(x) graph are (-4,4) and (4,-4)

-------------------------------------

Here is what i said before:

Part a y=f(2a)

f(-4)=4 but f(2a)=-4 so 2a=-4 so a=-2

f(4)=-4 but f(2b)=4 so 2b=4 so b=2

----------------------------------------

But in the next post I will do an entirley different example, perhaps you will be able to see how many graphs are related to parent graphs. In this case f(x) is the parent function and f(2x) it the 'offspring' Multiplying the x by 2 will cause the graph to have half the width (from the y axis).

Part (b): Find the graph of y=2x Verify that your points from part (a) are on the graph.

Part (c): The points (c,4) and (d,-4) are on the graph of y=f(2x-8) Find c and d

Part (d): Find the graph of y=f(2x-8) Be sure to verify that your points from part (c) are on the graph both algebraically and geometrically.

Melody
Feb 15, 2018

#3**0 **

Here is a different example:

let

\(f(x)=x^2\)

This is a concave up parabola with vertex (0,0)

Now \(f(2x) = (2x)^2\\f(2x) = 4x^2\)

How does the second one differ from the first.

Here is the graphs

The green one is f(x) and the blue one is f(2x)

Can you see that for any given y value the new corresponding x is half the size.

So when y=36 the x value was originally \(\pm6\) \(If \;f(x)=x^2 \;\;then \;\;f(6)=36\)

But when x is replaced with 2x then it becomes \( f(2x) =4x^2\) as show above and \(f(3)=36\)

**so if you double the x in the equation then the resulting graph will have half the width!**

Melody
Feb 15, 2018