How would you graph $y = ln (x - 1) + 3$ $y = ln(x-1) +3$ without a calculor?

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Answer 1 · 2016-09-30T02:01:42

This is the standard graph of $ln x$ $lnx$ shifted $+ 1$ $+1$ unit on the $x$ $x$ -axis and $+ 3$ $+3$ units on the $y$ $y$ -axis

Consider the standard function $ln x$ $lnx$
This can be considered as: $ln (x - a) + b$ $ln(x-a)+b$ where $a = b = 0$ $a=b=0$

The given function $y = ln (x - 1) + 3$ $y=ln(x-1)+3$ is the standard function with two linear transformations: $a = 1$ $a=1$ and $b = 3$ $b=3$

Hence: $y$ $y$ is the standard graph of $ln x$ $lnx$ shifted $+ 1$ $+1$ unit on the $x$ $x$ -axis and $+ 3$ $+3$ units on the $y$ $y$ -axis

This can be seen the graph of $y = ln (x - 1) + 3$ $y=ln(x-1)+3$ below:

graph{ln(x-1)+3 [-10, 10, -5, 5]}

How would you graph y = ln(x-1) +3y=ln(x−1)+3y = ln(x-1) +3 without a calculor?