Question

How do you find the slope of the secant lines of $f\left( x \right) = \left( {\dfrac{1}{x}} \right)$ through the points: $( - 4,\left( {f\left( { - 4} \right)} \right)$ and $\left( {1,f\left( 1 \right)} \right)$?

Answer 1

Hint: The equation of a straight line in slope-intercept form is: $y = mx + b$. Where m is the value of slope and b is the y-intercept. Here, m and b are constants, and x and y are variables. Since x and y are variables that describe the position of specific points on the graph, m and b describe features of the function. A straight line is a linear equation of the first order. The slope of a line is the ratio of change in y over the change in x between any two points on the line.
$slope\left( m \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
Steps to follow:
Find the slope of the line.
Use the slope to find the y-intercept.
Substitute the value of slope and y-intercept in a straight-line equation.

Complete step-by-step answer:
Here, we want to find a line equation. For that two points are given. Y-coordinate of the point is in the form of a function.
First, let us find the function value.
The given function is:
$ \Rightarrow f\left( x \right) = \left( {\dfrac{1}{x}} \right)$
Let us put the value of x is -4.
So,
$ \Rightarrow f\left( { - 4} \right) = \left( {\dfrac{1}{{ - 4}}} \right)$
That is equal to,
$ \Rightarrow f\left( { - 4} \right) = - \dfrac{1}{4}$
Now, let us put the value of x is 1.
So,
$ \Rightarrow f\left( 1 \right) = \dfrac{1}{1}$
That is equal to,
$ \Rightarrow f\left( 1 \right) = 1$
Hence, the two points are $\left( { - 4, - \dfrac{1}{4}} \right)$ and $\left( {1,1} \right)$.
Let us compare points $\left( { - 4, - \dfrac{1}{4}} \right)$ and $\left( {1,1} \right)$ with $\left( {{x_1},{y_1}} \right)$and $\left( {{x_2},{y_2}} \right)$
Therefore, ${x_1} = - 4,{y_1} = - \dfrac{1}{4}$ and ${x_2} = 1,{y_2} = 1$
Now, the first step is to find the slope.
$ \Rightarrow slope\left( m \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
Let us substitute all the values.
 $ \Rightarrow m = \dfrac{{ - \dfrac{1}{4} - 1}}{{ - 4 - 1}}$
Let us take the LCM of the numerator.
$ \Rightarrow m = \dfrac{{\dfrac{{ - 1 - 4}}{4}}}{{ - 4 - 1}}$
That is equal to,
$ \Rightarrow m = \dfrac{{\dfrac{{ - 5}}{4}}}{{ - 5}}$
$ \Rightarrow m = \dfrac{1}{4}$
Now, we will use the point-slope formula to find the equation for the line passing through these two points.
The point-slope formula is:
$ \Rightarrow \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$
Here, m is the slope and $\left( {{x_1},{y_1}} \right)$ is a point the line passes through.
Let us substitute these values in the above equation.
$ \Rightarrow \left( {y - \left( { - \dfrac{1}{4}} \right)} \right) = \dfrac{1}{4}\left( {x - \left( { - 4} \right)} \right)$
First, remove the brackets.
$ \Rightarrow \left( {y + \dfrac{1}{4}} \right) = \dfrac{1}{4}\left( {x + 4} \right)$
Multiply $\dfrac{1}{4}$ on the right-hand side.
$ \Rightarrow \left( {y + \dfrac{1}{4}} \right) = \dfrac{1}{4}x + \dfrac{1}{4}\left( 4 \right)$
That is equal to,
 $ \Rightarrow y + \dfrac{1}{4} = \dfrac{x}{4} + 1$
Let us subtract $\dfrac{1}{4}$ on both sides.
$ \Rightarrow y + \dfrac{1}{4} - \dfrac{1}{4} = \dfrac{x}{4} + 1 - \dfrac{1}{4}$
Let us simplify it.
$ \Rightarrow y = \dfrac{x}{4} + \dfrac{{4 - 1}}{4}$
That is equal to,
$ \Rightarrow y = \dfrac{1}{4}x + \dfrac{3}{4}$
Hence, the slope is $\dfrac{1}{4}$, and the equation of the line is$y = \dfrac{1}{4}x + \dfrac{3}{4}$.

Note:
We can find the line equation by selecting the second point that is $\left( {1,1} \right)$.
The point-slope formula is:
$ \Rightarrow \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$
Here, m is the slope and $\left( {{x_1},{y_1}} \right)$ is a point the line passes through.
Let us substitute these values in the above equation.
$ \Rightarrow \left( {y - 1} \right) = \dfrac{1}{4}\left( {x - 1} \right)$
First, remove the brackets.
$ \Rightarrow y - 1 = \dfrac{1}{4}x - \dfrac{1}{4}$
Let us subtract 1 on both sides.
 $ \Rightarrow y - 1 + 1 = \dfrac{1}{4}x - \dfrac{1}{4} + 1$
That is equal to,
$ \Rightarrow y = \dfrac{1}{4}x + \dfrac{{ - 1 + 4}}{4}$
So,
$ \Rightarrow y = \dfrac{1}{4}x + \dfrac{3}{4}$