Transformation of functions: "The point (1,-2) is on the graph of f(x)..."

[Don't Drink and Derive]

The point (1,-2) is on the graph of f(x). Describe the following transformations on f(x), and determine the resulting point.

So f(x) is starting at 0,0 I am guessing therefore f(x)=0. Then I add the points (1,-2) so f(x)=(x-1)-2

Then I input it into my three questions.

a) g(x)=2f(x)+3 inputting f(x) results in g(x)=2((x-1)-2)+3

b)g(x)=f(x+1)-3 inputting f(x) results in g(x)=(((x-1)-2)+1)-3 simplified is g(x)=((x-1)-1)-3

c)g(x)=-f(2x) inputting f(x) results in g(x)=-2((x-1)-2)

d)g(x)=-f(-x-1)+3 inputting results in g(x)=-(-((x-1)-2)-1)+3 simplified (((x-1)-2)+1)+3 g(x)=((x-1)-1)+3

Much appreciated

 

[Don't Drink and Derive]

Trying to keep all the answers in y=af(k(x-d)+h)+c

 

[Harry_the_cat]

I don't think you are understanding the question.
You are told that (1, -2) is on the graph of y=f(x). (You can't assume anything about (0, 0) and don't need to.)

You first need to understand what various parameters "transform" the graph of a function.

The way that the graph of y = g(x) = a*f(b(x-c))+d transforms the graph of y = f(x) depends on the values of a, b, c and d.

Do you know the effect of each of these parameters?

Let's look at a) g(x) = 2*f(x) + 3

The a=2 multiplies the f(x) or y-value by 2 (In general "a" stretches or compresses the graph vertically, or reflects in in x-axis if a is neg).
The d=+3 adds 3 to the y-value (In general, +d moves the graph up d units, -d moves it down).

So in this case, the graph of y=f(x) is being stretched vertically by a factor of 2 and then moved up 3 units.

Every point on the graph of y=f(x) is transformed in this way.

So (1, -2) moves to (1, -4) after being stretched 2x vertically, then (1, -4) moves to (1, -1) after being moved up 3 units.
So the resulting point is (1, -1).

That's what is happening graphically.

Algebraically:
If (1, -2) is a point on y=f(x) then f(1) = -2

Now g(x) = 2*f(x) +3 so g(1) = 2*f(1) + 3 = 2*(-2)+3 =-4 + 3 = -1 resulting in the point (1, -1)

 

[Don't Drink and Derive]

*face palm* less complicated than I thought

So b) f(x) is horizontally translated 1 unit to the right and vertically translated 3 units down. Therefore new point is (0,-5)

Correction (2,-5)

Thanks for clearing things up!

 

[Don't Drink and Derive]

*face palm* less complicated than I thought

So b) f(x) is horizontally translated 1 unit to the right and vertically translated 3 units down. Therefore new point is (0,-5)

Correction (2,-5)

Thanks for clearing things up!

Algebraic
G(x)=f(x+1)-3
G(1)=f((1)+1)-3
G(1)=(-2+1)-3
G(1)=-1-3
G(1)=-4

So the new y is -4 and input this back in and solve for x when y is -4?

-4=(x+1)-3
-7=x+1
-6=x

New point is (-6,-4)

 

[Harry_the_cat]

*face palm* less complicated than I thought

So b) f(x) is horizontally translated 1 unit to the right and vertically translated 3 units down. Therefore new point is (0,-5)

Correction (2,-5)

Thanks for clearing things up!

I unit to the LEFT and 3 down. (0, -5)

 

[Harry_the_cat]

Algebraic
G(x)=f(x+1)-3
G(1)=f((1)+1)-3
G(1)=(-2+1)-3
G(1)=-1-3
G(1)=-4

So the new y is -4 and input this back in and solve for x when y is -4?

-4=(x+1)-3
-7=x+1
-6=x

New point is (-6,-4)

It gets messy when you try to do it using algebra. Best to think what happens to the graph.

 

[Don't Drink and Derive]

It gets messy when you try to do it using algebra. Best to think what happens to the graph.

Will do, and last thing is for c it's a reflection about the x axis and a horizontal compression by a factor of 1/2. I got it now
Cheers