trying to help my daughter with this problem

Question

) Let f(x) = e x – 2  + 5.                                                                                                                   

 

Answers can be stated without additional work/explanation.

 

      (a) Which describes how the graph of f can be obtained from the graph of y = ex ?    Choice: ________

 

            A.     Shift the graph of  y = ex  to the left by 2 unit and up by 5 units.

            B.     Shift the graph of  y = ex  to the right by 2 unit and up by 5 units.

            C.     Reflect the graph of  y = ex  across the x-axis and shift up by 5 units.

            D.     Reflect the graph of  y = ex  across the y-axis and shift up by 5 units.

 

      (b) What is the domain of f ?

 

 

 

      (c) What is the range of f ?

 

 

 

      (d) What is the horizontal asymptote?

 

 

 

      (e) What is the y-intercept?  State the approximation to 2 decimal places (i.e., the nearest hundredth).

Answer 1

Hi Reginald, the function: e^x-2 + 5, has the following qualities:

 

The function is shifted to the right 2 units and up 5 units.

 

The domain of x is: -∞ to ∞.

 

The range of y is: 5 to ∞.

 

The horizontal asymptote is: y = 5.

 

The y-intercept is: approximately 5.14

 

Questions?

Answer 2

(a) B.

(b) ℜ

(c) f(x) > 5

(d) f(x) = 5

(e) f(0) = e^-2 + 5, 5.14

Answer 3

Your problem statement is incorrect.  From clues in the answers to (a), (x-2) has to be an exponent.

When typing, you can use the symbol ^ to indicate raising to a power (i.e. an exponent).

Also parentheses are critically important.

 

f(x) = e^(x-2) + 5

 

a) The answer is B.  An adjustment directly to x shifts the graph left for positive and right for a negative adder.

     An addition of a constant (the +5 at the end) adds to the y value.  Positive shifts the graph up; negative shifts it down.

 

b) The domain is all real numbers.  No real value of x causes the calculation to be undefined.

 

c) The range of the function is (5, ∞), or  5<y<∞.  Note that the 5 itself is NOT included in the range.

 

d) The graph approaches, but never reaches, 5 as x becomes more negative.

     The horizontal asymptote is y=5.

 

e) A y-intercept always occurs when x=0.

     use a calculator

     f(x) = e^(-2) + 5 = 5.14