Stefan said that the y-intercept for the graph of the function f(x)=3log(x−7)+1 is 1. He also said that as x approaches positive infinity, the graph of f(x) approaches positive infinity. Describe which of Stefan's statements, if any, are incorrect, and, if so, correct the statements.

see below

Step-by-step explanation:

f(x)=3log(x−7)+1

There is no y intercept

log (x) does not intercept the y axis.  This shifts the function 7 units to the right so the graph will not intercept the y axis

Answers:

His first statement is incorrect

His second statement is correct.

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Explanation:

The first statement is incorrect because plugging x = 0 into f(x) leads to

f(x) = 3*log(x-7)+1

f(0) = 3*log(0-7)+1

f(0) = 3*log(-7)+1

We stop here because we cannot take the log of a negative number (unless you want a complex number but that's for another time). This means that x = 0 is not in the domain. The domain is x > 7 to ensure that x-7 > 0. That way we avoid taking the log of a negative number, and taking the log of zero as well.

The vertical asymptote is at x = 7 and the log curve is entirely to the right of it, so there's no way it's reaching the y axis.

Simply put: there is no y intercept for this function.

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The second statement is correct.

The function y = log(x) has y go off to infinity when x does.

The same applies to y = 3log(x) and y = 3log(x-7) and y = 3log(x-7)+1

We can say it "rises to the right" as another way of describing part of the end behavior.

A graph can confirm this as shown below.

the answer is d. (1.2, -4.7) on edgn

hope this !

idk

step-by-step explanation: