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Question: How do you find the slope of the secant lines of \(f\left( x \right) = \left( {\dfrac{1}{x}} \right)...

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

Explanation

Solution

The equation of a straight line in slope-intercept form is: y=mx+by = 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(m)=y2y1x2x1slope\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:
f(x)=(1x)\Rightarrow f\left( x \right) = \left( {\dfrac{1}{x}} \right)
Let us put the value of x is -4.
So,
f(4)=(14)\Rightarrow f\left( { - 4} \right) = \left( {\dfrac{1}{{ - 4}}} \right)
That is equal to,
f(4)=14\Rightarrow f\left( { - 4} \right) = - \dfrac{1}{4}
Now, let us put the value of x is 1.
So,
f(1)=11\Rightarrow f\left( 1 \right) = \dfrac{1}{1}
That is equal to,
f(1)=1\Rightarrow f\left( 1 \right) = 1
Hence, the two points are (4,14)\left( { - 4, - \dfrac{1}{4}} \right) and (1,1)\left( {1,1} \right).
Let us compare points (4,14)\left( { - 4, - \dfrac{1}{4}} \right) and (1,1)\left( {1,1} \right) with (x1,y1)\left( {{x_1},{y_1}} \right)and (x2,y2)\left( {{x_2},{y_2}} \right)
Therefore, x1=4,y1=14{x_1} = - 4,{y_1} = - \dfrac{1}{4} and x2=1,y2=1{x_2} = 1,{y_2} = 1
Now, the first step is to find the slope.
slope(m)=y2y1x2x1\Rightarrow slope\left( m \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}
Let us substitute all the values.
m=14141\Rightarrow m = \dfrac{{ - \dfrac{1}{4} - 1}}{{ - 4 - 1}}
Let us take the LCM of the numerator.
m=14441\Rightarrow m = \dfrac{{\dfrac{{ - 1 - 4}}{4}}}{{ - 4 - 1}}
That is equal to,
m=545\Rightarrow m = \dfrac{{\dfrac{{ - 5}}{4}}}{{ - 5}}
m=14\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:
(yy1)=m(xx1)\Rightarrow \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)
Here, m is the slope and (x1,y1)\left( {{x_1},{y_1}} \right) is a point the line passes through.
Let us substitute these values in the above equation.
(y(14))=14(x(4))\Rightarrow \left( {y - \left( { - \dfrac{1}{4}} \right)} \right) = \dfrac{1}{4}\left( {x - \left( { - 4} \right)} \right)
First, remove the brackets.
(y+14)=14(x+4)\Rightarrow \left( {y + \dfrac{1}{4}} \right) = \dfrac{1}{4}\left( {x + 4} \right)
Multiply 14\dfrac{1}{4} on the right-hand side.
(y+14)=14x+14(4)\Rightarrow \left( {y + \dfrac{1}{4}} \right) = \dfrac{1}{4}x + \dfrac{1}{4}\left( 4 \right)
That is equal to,
y+14=x4+1\Rightarrow y + \dfrac{1}{4} = \dfrac{x}{4} + 1
Let us subtract 14\dfrac{1}{4} on both sides.
y+1414=x4+114\Rightarrow y + \dfrac{1}{4} - \dfrac{1}{4} = \dfrac{x}{4} + 1 - \dfrac{1}{4}
Let us simplify it.
y=x4+414\Rightarrow y = \dfrac{x}{4} + \dfrac{{4 - 1}}{4}
That is equal to,
y=14x+34\Rightarrow y = \dfrac{1}{4}x + \dfrac{3}{4}
Hence, the slope is 14\dfrac{1}{4}, and the equation of the line isy=14x+34y = \dfrac{1}{4}x + \dfrac{3}{4}.

Note:
We can find the line equation by selecting the second point that is (1,1)\left( {1,1} \right).
The point-slope formula is:
(yy1)=m(xx1)\Rightarrow \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)
Here, m is the slope and (x1,y1)\left( {{x_1},{y_1}} \right) is a point the line passes through.
Let us substitute these values in the above equation.
(y1)=14(x1)\Rightarrow \left( {y - 1} \right) = \dfrac{1}{4}\left( {x - 1} \right)
First, remove the brackets.
y1=14x14\Rightarrow y - 1 = \dfrac{1}{4}x - \dfrac{1}{4}
Let us subtract 1 on both sides.
y1+1=14x14+1\Rightarrow y - 1 + 1 = \dfrac{1}{4}x - \dfrac{1}{4} + 1
That is equal to,
y=14x+1+44\Rightarrow y = \dfrac{1}{4}x + \dfrac{{ - 1 + 4}}{4}
So,
y=14x+34\Rightarrow y = \dfrac{1}{4}x + \dfrac{3}{4}