Can someone help me solve these last 4 problems..

Question

A) y=5-e^-x B) y=e^(x-2) C) y=log(2) X D) y=3+log(2) X?

find : -horizontal and vertical asymptotes. -x and y intercepts -domain and range -graph and label two points

Answer 1

(a)

The Function is y = 5 - e^(-x)

(1)

Domain of the Function y = 5 - e^(-x) is set of all real numbers.

Domain : {x Є R}.

(2)

Range of the Function is y < 5.

(3)

To find the Horizontal Asymptotes, We apply limit at x tends to +∞ and -∞.

We do not consider limit at x tends to -∞ because it is no

The Horizontal Asymptotes of y = 5 - e^(-x) is 5

(4)

There are no Vertical Asymptotes of f(x)

(5)

To find the x intercept, we substitute y = 0 in the Equation.

5 - e^(-x) = 0

e^(-x) = 5

- x = 1.609        (Using Logarithmic table ln(5) = 1.609)

x = -1.609

Therefore x intercept is ( -1.609,0)

(5)

To find the y intercept, we substitute x= 0 in the Equation.

5 - e^(-0) = y

y = 5 - 1

y = 4

Therefore y intercept is (0,4)

(6)

Graph

Therefore

Domain of y = 5 - e^(-x) is {x Є R}.

Range of the Function is y < 5.

The Horizontal Asymptotes of y = 5 - e^(-x) is 5.

x intercept is ( -1.609,0).

y intercept is (0,4).

Answer 2

(b)

The Function is y = e^(x-2)

(1)

Domain of the Function y = e^(x-2) is set of all real numbers.

Domain : {x Є R}.

(2)

For all Values of x the value of y is positive.

Range of the Function is y > 0.

(3)

To find the Horizontal Asymptotes, We apply limit at x tends to +∞ and -∞.

=

= 0

We do not consider limit at x tends to ∞ because it is not defined

Therefore the Horizontal Asymptote is x = 0.

(4)

There are no Vertical Asymptotes of function y = e^(x-2)

(5)

To find the x intercept, we substitute y = 0 in the Equation.

e^(x-2) = 0

Exponent function cannot be equated to zero.

There are no x intercept for the Function y = e^(x-2)

(6)

To find the y intercept, we substitute x= 0 in the Equation.

e^(0-2) = y

y = 0.1353

Therefore y intercept is (0,0.1353)

(7)

Graph

Therefore

Domain of y = e^(x-2) is {x Є R}.

Range of the Function is y > 2.

The Horizontal Asymptotes of y = e^(x-2) is 0.

There are No vertical Asymptotes of y = e^(x-2)

There are no x-intercept.

y intercept is (0,0.1353).

Answer 3

(c)

The Function is .

(1)

Domain of the Function is set of positive number.

Domain : {x:x>0}.

(2)

For all Values of x the value of y is exist.

Range of the Function is R

(3)

There are no Horizantal Asymptotes of function

(4)

The Function is

The vertical asymptote is obtained solving x = 0

Therefore the vertical asymptote is x = 0.

(5)

To find the x intercept, we substitute y = 0 in the Equation.

log(x) / log2 = 0

Apply Exponential Function on both sides.

x = 1

Therefore x intercept of is x = 1.

(6)

To find the y intercept, we substitute x= 0 in the Equation.

log(0)/log(2) = y

log(0) is not defined.

Therefore there is no y intercept for the function

(7)

Graph

Therefore

Domain of is {x:x>0}.

Range of the Function is R.

There is no Horizontal Asymptotes of .

The is vertical Asymptotes of is x = 0.

The x-intercept of is x =1.

There is no y-intercept.

 

Answer 4

(D)

The Function is .

(1)

Domain of the Function is set of positive number.

Domain : {x:x>0}.

(2)

For all Values of x the value of y is exist.

Range of the Function is R.

(3)

There are no Horizantal Asymptotes of function

(4)

The Function is

The vertical asymptote is obtained solving x = 0

Therefore the vertical asymptote is x = 0.

(5)

To find the x intercept, we substitute y = 0 in the Equation.

3 +log(x) / log2 = 0

logx = - 3 * log2

logx = -2.079

Apply Exponential Function on both sides.

x = 0.125

Therefore x intercept of is x = 0.125.

(6)

To find the y intercept, we substitute x= 0 in the Equation.

3+log(0)/log(2) = y

log(0) is not defined.

Therefore there is no y intercept for the function

(7)

Graph

Therefore

Domain of is {x:x>0}.

Range of the Function is R.

There is no Horizontal Asymptotes of .

The is vertical Asymptotes of is x = 0.

The x-intercept of is x =0.125.

There is no y-intercept.