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Question: How would you graph \( y = \ln \left( {x - 1} \right) + 3 \) ?...

How would you graph y=ln(x1)+3y = \ln \left( {x - 1} \right) + 3 ?

Explanation

Solution

Hint : A graph of a function f is the set of ordered pairs; the equation of graph is generally represented as y=f(x)y = f\left( x \right) , where x and f(x)f\left( x \right) are real numbers. We substitute the value of x and we determine the value of y and then we mark the points in the graph and we join the points.

Complete step by step solution:
Here, in the given question, we have to plot the graph for the given function. A graph of a function is a set of ordered pairs and it is represented as y=f(x)y = f\left( x \right) , where x and f(x)f\left( x \right) are real numbers. These pairs are in the form of cartesian coordinates and the graph is the two-dimensional graph.
First, we have to find the value of y by using the graph equation y=ln(x1)+3y = \ln \left( {x - 1} \right) + 3 .
Let us substitute the value of x as 22 .
y=ln(21)+3\Rightarrow y = \ln \left( {2 - 1} \right) + 3
y=ln(1)+3\Rightarrow y = \ln \left( 1 \right) + 3
We know that the value of ln(1)\ln \left( 1 \right) is zero. So, we get,
y=0+3\Rightarrow y = 0 + 3
y=3\Rightarrow y = 3
Now we consider the value of x as e+1e + 1 , the value of y is
y=ln((e+1)1)+3\Rightarrow y = \ln \left( {\left( {e + 1} \right) - 1} \right) + 3
Simplifying the expression,
y=ln(e)+3\Rightarrow y = \ln \left( e \right) + 3
Now, we know that the value of ln(e)\ln \left( e \right) is one. So, we get,
y=1+3\Rightarrow y = 1 + 3
y=4\Rightarrow y = 4
Now we consider the value of x as e2+1{e^2} + 1 , the value of y is
y=ln((e2+1)1)+3\Rightarrow y = \ln \left( {\left( {{e^2} + 1} \right) - 1} \right) + 3
Simplifying the expression,
y=ln(e2)+3\Rightarrow y = \ln \left( {{e^2}} \right) + 3
Now, we know that the value of ln(e)\ln \left( e \right) is one. So, we get,
y=2+3\Rightarrow y = 2 + 3
y=5\Rightarrow y = 5
Now we draw a table for these values we have

x22(e+1)\left( {e + 1} \right)(e2+1)\left( {{e^2} + 1} \right)
y334455

We also know the nature of the graph of logarithmic function. Hence, we can now plot the graph of the given function y=ln(x1)+3y = \ln \left( {x - 1} \right) + 3 with the help of coordinates of the points. The nature of the graph of a function and its slope can also be determined from the derivative of the function. The graph plotted for these points is represented below:

Note : The number ‘e’ in mathematics is known as the euler’s number. It is a mathematical constant and has a value approximately equal to 2.718282.71828 . It is the base of the natural logarithm that is used in many steps in the given problem.