Pre calc graphing help?

Question

How do you solve this 

Find the domain, range, asymptote of the following functions 

1). y= e^-3x. +1 


2). y=log (x+2)

Answer 1

1) The general form of an exponential function is defined by f (x ) = e ^x + k

where a  is positive constant , a  not equals to 1 and a is called base. x is any real number.

The graph has horizontal asympotote of y = k .

Horizontal asympotote y = 1

f (x ) = e^-3x + 1

y  = e^-3x + 1

We will find the domain, range and asymptote by looking at the graph of the exponential function.

 

Answer 2

Choose values for x and find the corresponding values for y.

x	y = e-3x + 1	(x, y )
2	y = e-3x + 1 = (2.71828)-3*2 + 1 = (2.71828)-6 + 1 = 1/403.4271+ 1 = 0.0024+ 1 = 1.0024	(2,1.0024)
1	y = e-3x + 1 = (2.71828)-3*1 + 1 = (2.71828)-3 + 1 = 1/20.0854+ 1 = 0.04978+ 1 = 1.04978	(1,1.04978)
0	y = e-3x + 1 = e-3*0 + 1 = e0 + 1 = 1+1=2	(0,2)
-0.1	y = e-3x + 1 = (2.71828)-3*-0.1 + 1 = (2.71828)0.3 + 1 = 1.34985+ 1 = 2.34985	(-0.1,2.34985)
-0.2	y = e-3x + 1 = (2.71828)-3*-0.2 + 1 = (2.71828)0.6 + 1 = 1.8221+ 1 = 2.8221	(-0.2,2.8221)
-0.5	y = e-3x + 1 = (2.71828)-3*-0.5 + 1 = (2.71828)1.5 + 1 = 4.48168+ 1 = 5.48168	(-0.5,5.48168)

 

Answer 3

1.Draw a coordinate plane.

2.Plot the coordinate points and dash the horizontal asympotote.

3.Then sketch the graph, connecting the points with a smooth curve.

The domain is all positive real numbers (never zero).

The range is (1, infinity)

Horizontal asymptote is y=1.

Answer 4

2) The general form of an exponential function is defined by f (x ) = log(x-k)

where a  is positive constant , a  not equals to 1 and a is called base. x is any real number.

The graph has asympotote of y = k .

Horizontal asympotote y = -2

f (x ) = log(x+2)

y  = log(x+2)

We will find the domain, range and asymptote by looking at the graph of the exponential function.

Choose values for x and find the corresponding values for y.

x	y = log(x+2)	(x, y )
-1	y = log(x+2) = log(-1+2) = log1=0	(-1,0)
-1.8	y = log(x+2) = log(-1.8+2) = log-0.2=0.6987	(-1.8,0.6987)
0	y = log(x+2) = log(0+2) = log2=0.3010	(0,0.3010)
1	y = log(x+2) = log(1+2) = log3=0.4771	(1,0.4771)
1.5	y = log(x+2) = log(1.5+2) = log3.5=0.5440	(1.5,5440)

1.Draw a coordinate plane.

2.Plot the coordinate points and dash the horizontal asympotote.

3.Then sketch the graph, connecting the points with a smooth curve.

The domain is x = all real numbers.

The domain is y = (-2, infinity).

Horizontal asymptote is y=-2.

 

 

Comments

Observe the graph,

Domain : {x ∈ R : x > - 2},

Range : R (all real numbers)

Vertical asymptote : x = - 2.