shifting and scaling

Question

Shift the graph of p(x)=1/x so that x=-9 is the vertical asymptote and y=2 is the horizontal asymptote of the shifted graph. Write the expression of the shifted function.

 

Shift the graph of h(x)=e^x so that y=-6.25 is the horizontal asymptote of the shifted graph. Write the expression of the shifted function

Answer 1

For the the first function to find the shift to get a vertical asymptote of x=-9 you have to find how to make the denominator of the function p(x) equal to zero when x=-9. The horizontal asymptote is found by taking the limit of the function as x=>±∞ so that the limit equals 2. The easiest way to do this is to add a constant to the expression. That way when the fraction goes to zero at infinity you are still left with a number that is not reliant on "x".The final expression should have the form of p(x)=(1/(x+a))+b where "a" and "b" are numbers.

 

To do this one you follow the same process for finding the horizontal asymptote for the previous problem except h(x)=e^x has two limits. lim e^x as x approaches infinity is infinity whereas when x approaches negative infinity it equals zero. So, for this shift you take the limit as e^x approaches negative infinity and add your constant to shift the graph down to -6.25. The final expression should look like h(x)=(e^x)+a where "a" is a constant.

 

 

Answer 2

If you can’t remember them, the translation rules are easily found by drawing two circles with the same radius, r, on the x-y coordinate plane; one centered at A(0,0) and the other centered at B(h,k).

The equations of each can separately be derived by using the Pythagorean Theorem with a point (x,y) on the circle.

The equations you get are:

A(0,0): x^2 + y^2 = r^2

B(h,k): (x-h)^2 + (y-k)^2 = r^2

Circle B is a translation of circle A by h units in the x direction and k units in the y direction (a translation vector of <h,k>):

[x^2 + y^2 = r^2] –> [(x-h)^2 + (y-k)^2 = r^2]

So the rule for equations is:

change every occurrence of x to (x – h)
and every occurrence of y to (y – k).

Since point A(0,0) changed to point B(h,k), the rule for every point is:

add h to the x-coordinate and k to the y-coordinate.

Let’s use these rules for this question.

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“Shift the graph of p(x) = 1/x so that x = -9 is the vertical asymptote and y = 2 is the horizontal asymptote of the shifted graph. Write the expression of the shifted function.”

p(x) = 1/x is a rational function with a vertical asymptote x = 0 and a horizontal asymptote y = 0. Most precalculus texts start their sections on rational functions with this function; so it, and its asymptotes, should be familiar.

The function we want has asymptotes x = -9, a left shift of 9 from p(x)’s x = 0, and y = 2, a shift up of 2 from p(x)’s y = 0. So <h,k> = <-9,2> and if p –> q then:

q(x) – 2 = p(x – (-9)) = p(x + 9) = 1/(x + 9), or

q(x) = 1/(x + 9) + 2.

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“Shift the graph of h(x) = e^x so that y=-6.25 is the horizontal asymptote of the shifted graph. Write the expression of the shifted function”

As developed in precalculus texts at the beginning of the exponential functions section, all functions of the form f(x) = a b^x, a ≠ 0, b > 0, b ≠ 1, have a horizontal asymptote of y = 0.

The function we want has a horizontal asymptote of y = -6.25, a shift down of 6.25 from h(x)’s y = 0. So <h,k> = <0,-6.25> and if h –> g then:

g(x) – (-6.25) = h(x – 0) = e^x, or

g(x) = e^x – 6.25

Answer 3

deleted duplicate answer