Chapter 12.3, Problem 58E

To determine

ToFind:The equation of the line that is tangent to the graph of $f\left(x\right)=x^2-x$ and parallel to the line $x+2y-6=0$

Answer to Problem 58E

The equation of the tangent line to the graph of $f\left(x\right)=x^2-x$ and parallel to the line $x+2y-6=0$ is $y=-\frac{1}{2}x-\frac{1}{16}$

Explanation of Solution

Given:

The function $f\left(x\right)=x^2-x$ and the line $x+2y-6=0$

Concept Used:

The slope $m$ of the tangent lineto the graph of $f$ at $\left(x,f\left(x\right)\right)$ , and is given by

  $m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

The equation of the tangent line to the graph of a function $f$ at the point $\left(a,b\right)$ with slope $m$ is given by

  $y-b=m\left(x-a\right)$

  ${\left(a+b\right)}^2=a^2+b^2+2ab$

Calculation:

Forthe function $f\left(x\right)=x^2-x$

The slope $m$ of the tangent line to the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ is given by

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{\left({\left(x+h\right)}^2-\left(x+h\right)\right)-\left(x^2-x\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{x^2+h^2+2xh-x-h-x^2+x}{h}\right){\left(a+b\right)}^2=a^2+b^2+2ab\end{array}$

  $\begin{array}{c}=\underset{h\to 0}{\lim }\left(\frac{h^2+2xh-h}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{h\left(h+2x-1\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(h+2x-1\right)\\ =0+2x-1\text{ Substituting limit}\\ =2x-1\end{array}$

Consider the line $x+2y-6=0$

Writing the line $x+2y-6=0$ into slope intercept form

  $\begin{array}{c}2y=-x+6\\ y=-\frac{1}{2}x+3\end{array}$

The slope of the line $x+2y-6=0$ is $-\frac{1}{2}$

Since, the tangent line is parallel to the line $x+2y-6=0$ . So, their slopes will be equal.

  $\begin{array}{c}2x-1=-\frac{1}{2}\\ 2x=-\frac{1}{2}+1\\ 2x=\frac{1}{2}\\ x=\frac{1}{4}\end{array}$

Substituting in the function $y=f\left(x\right)=x^2-x$

  $\begin{array}{c}y={\left(\frac{1}{4}\right)}^2-\frac{1}{4}\\ =\frac{1}{16}-\frac{1}{4}\\ =-\frac{3}{16}\end{array}$

So, a point on the tangent line is $\left(x,y\right)=\left(\frac{1}{4},-\frac{3}{16}\right)$

Also, the slope of tangent line is $m=-\frac{1}{2}$

The equation of the tangent line at the point $\left(a,b\right)$ is given by

  $y-b=m\left(x-a\right)$

  $\begin{array}{c}\left(y-\left(-\frac{3}{16}\right)\right)=-\frac{1}{2}\left(x-\frac{1}{4}\right)\\ \left(y+\frac{3}{16}\right)=-\frac{1}{2}\left(x-\frac{1}{4}\right)\\ 2\left(y+\frac{3}{16}\right)=-1\left(x-\frac{1}{4}\right)\\ 2y+\frac{3}{8}=-x+\frac{1}{4}\end{array}$

  $\begin{array}{c}2y=-x+\frac{1}{4}-\frac{3}{8}\\ 2y=-x-\frac{1}{8}\\ y=-\frac{1}{2}x-\frac{1}{16}\end{array}$

Therefore, the equation of the tangent line to the graph of $f\left(x\right)=x^2-x$ and parallel to the line $x+2y-6=0$ is $y=-\frac{1}{2}x-\frac{1}{16}$

Conclusion:

Theequation of the tangent line to the graph of $f\left(x\right)=x^2-x$ and parallel to the line $x+2y-6=0$ is $y=-\frac{1}{2}x-\frac{1}{16}$