Which function has an asymptote at x= 39
2 answers:
Answer:
option B :
Step-by-step explanation:
(a)
For exponential function , there is no vertical asymptotes
General form of exponential function is
In the given f(x) the value of k =0
So horizontal asymptote is y=0
(b) lets check with option
To find vertical asymptote we set the argument of log =0 and solve for x
Argument of log is x-39
x-39=0 so x=39
Hence vertical asymptote at x=39
You might be interested in
Answer:
We conclude that the initial value and y-intercept are the same thing on a graph.
Please check the attached graph of the equation y = 2x+1.
Step-by-step explanation:
We know that the initial value on a graph is basically the out-put value y of the point where the line meets or crosses the y-axis.
In other words, the initial value is the y-value or output of the point at x = 0
For example,
Let the equation
y = 2x+1
substitute x = 0
y = 2(0)+1
y = 0+1
y = 1
Thus, the initial value of the equation y = 2x+1 is: y = 1
Please check the attached graph of the equation y = 2x+1.
It is clear from the graph that at x = 0, the value of y = 1.
Thus, at y = 1, the line meets the y-axis.
Hence, the initial value of the line is: y = 1
Similarly, we know that the value of the y-intercept can be determined by setting x = 0 and determining the corresponding value of y.
For example,
Let the equation
y = 2x+1
substitute x = 0
y = 2(0)+1
y = 0+1
y = 1
Thus, the y-intercept of y = 2x+1 is y = 1.
Please check the attached graph of the equation y = 2x+1.
It is clear from the graph that at x = 0, the value of y = 1.
Therefore, the y-intercept of y = 2x+1 is y = 1.
Conclusion:
Therefore, we conclude that the initial value and y-intercept are the same thing on a graph.
Please check the attached graph of the equation y = 2x+1.
