Finding the Equations of Reciprocal Function Graph

[MapleSpy]

The general equation for transformations is:
a f (k (x-d) ) + c

By looking at the vertical asymptote and horizontal asymptote you can find the values of d and c. How would I find the value of a also known as the vertical stretch/compression factor? To find the horizontal compression/stretch factor would I sub a point from the graph to find the value of k? For example... how would I find the value of a and k from the graph attached?

 

[stapel]

The general equation for transformations is:
a f (k (x-d) ) + c

By looking at the vertical asymptote and horizontal asymptote you can find the values of d and c. How would I find the value of a also known as the vertical stretch/compression factor? To find the horizontal compression/stretch factor would I sub a point from the graph to find the value of k? For example... how would I find the value of a and k from the graph attached?

Not knowing the function with which they'd started, I see no means of determining how it might have been transformed. Sorry.

 

[lookagain]

The general equation for transformations is:
a f (k (x-d) ) + c

By looking at the vertical asymptote and horizontal asymptote you can find the values of d and c. How would I find the value of a also known as the vertical stretch/compression factor? . . .

With [my] experience, the equation of the likely parent function is \(\displaystyle \ \ f(x) \ = \ \dfrac{1}{x}.\)

If that is the case, that parent function is symmetric with respect to the origin, and it has a horizontal asymptote of y = 0 and a vertical asymptote of x = 0.

Then that parent function was reflected across the x-axis, or equivalently, across the y-axis, and it had a horizontal shift 3 units to the right.

Look at a convenient point on the graph in the diagram just to the right of the vertical asymptote, in particular, (4, -2). Imagine reflecting it across the x-axis and shifting it the left 3 units.

You will get the point (1, 2). In the parent graph, that would have corresponded to the point (1, 1). What vertical stretch (or was it a compression?) factor would that make it to go from the

parent function to the function in the diagram? Work on these hints/suggestions, please.

 

[MapleSpy]

With [my] experience, the equation of the likely parent function is \(\displaystyle \ \ f(x) \ = \ \dfrac{1}{x}.\)

If that is the case, that parent function is symmetric with respect to the origin, and it has a horizontal asymptote of y = 0 and a vertical asymptote of x = 0.

Then that parent function was reflected across the x-axis, or equivalently, across the y-axis, and it had a horizontal shift 3 units to the right.

Look at a convenient point on the graph in the diagram just to the right of the vertical asymptote, in particular, (4, -2). Imagine reflecting it across the x-axis and shifting it the left 3 units.

You will get the point (1, 2). In the parent graph, that would have corresponded to the point (1, 1). What vertical stretch (or was it a compression?) factor would that make it to go from the

parent function to the function in the diagram? Work on these hints/suggestions, please.

Sent from my iPhone using Tapatalk

 

[MapleSpy]

So the vertical stretch factor is 2?


Sent from my iPhone using Tapatalk