Chapter 12.3, Problem 56E

To determine

ToFind:The equation of the line that is tangent to the graph of $f\left(x\right)=x^2+1$ and parallel to the line $2x+y=0$

Answer to Problem 56E

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

Explanation of Solution

Given:

The function $f\left(x\right)=x^2+1$ and the line $2x+y=0$

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 $f\left(x\right)=x^2+1$

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

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{\left({\left(x+h\right)}^2+1\right)-\left({\left(x\right)}^2+1\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{x^2+h^2+2xh+1-x^2-1}{h}\right){\left(a+b\right)}^2=a^2+b^2+2ab\end{array}$

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

  $=2x$

Consider the line $2x+y=0$

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

  $y=-2x$

The slope of the line $2x+y=0$ is $-2$

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

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

Substituting in the function $y=f\left(x\right)=x^2+1$

  $\begin{array}{c}y={\left(-1\right)}^2+1\\ =1+1\\ =2\end{array}$

So, a point on the tangent line is $\left(x,y\right)=\left(-1,2\right)$

Also, the slope of tangent line is $m=-2$

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

  $y-b=m\left(x-a\right)$

  $\begin{array}{c}\left(y-2\right)=-2\left(x-\left(-1\right)\right)\\ y-2=-2x-2\\ y=-2x-2+2\\ y=-2x\end{array}$

  $2x+y=0$

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

Conclusion:

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