Chapter 11.3, Problem 45E

a.

To determine

To calculate: The slope of the function, $f\left(x\right)=x^2+5x+6$ at the point $\left(-4,2\right)$ .

Answer to Problem 45E

The slope of the function, $f\left(x\right)=x^2+5x+6$ at the point $\left(-4,2\right)$ is -3.

Explanation of Solution

Given information: The function and point are $f\left(x\right)=x^2+5x+6$ and $\left(-4,2\right)$ respectively.

Formula used: The slope of the function $f\left(x\right)$ at the point $\left(x,f\left(x\right)\right)$ is given by $m_{\sec }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ such that the limit exists.

Calculation: Find the slope of the function, $f\left(x\right)=x^2+5x+6$ at the point $\left(-4,2\right)$ .

  $\begin{array}{c}{m}_{\sec }=\underset{h\to 0}{\lim }\frac{f\left(-4+h\right)-f\left(-4\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left[{\left(-4+h\right)}^2+5\left(-4+h\right)+6\right]-\left[{\left(-4\right)}^2+5\left(-4\right)+6\right]}{h}\\ =\underset{h\to 0}{\lim }\frac{\left[16+h^2-8h-20+5h+6\right]-\left[16-20+6\right]}{h}\\ =\underset{h\to 0}{\lim }\frac{2+h^2-3h-2}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(h-3\right)}{h}\\ =\underset{h\to 0}{\lim }\left(h-3\right)\end{array}$

Find the value of the limit when h approaches to 0.

  $\begin{array}{c}{m}_{\sec }=0-3\\ =-3\end{array}$

Thus, the slope of the function is -3.

b.

To determine

To calculate: The equation of a tangent line of the function, $f\left(x\right)=x^2+5x+6$ passing through $\left(-4,2\right)$ .

Answer to Problem 45E

The equation of a tangent line of the function, $f\left(x\right)=x^2+5x+6$ passing through $\left(-4,2\right)$ is $y=-3x-10$ .

Explanation of Solution

Given: The function and point are $f\left(x\right)=x^2+5x+6$ and $\left(-4,2\right)$ respectively.

Formula used: An equation of the line with slope $m$ and passing through the point $\left(x_1,y_1\right)$ is given by $y-y_1=m\left(x-x_1\right)$ .

Calculation: As derived in part a, the slope of the function is -3.

Find the equation of tangent line with slope -3 and passing through $\left(-4,2\right)$ .

  $\begin{array}{c}y-2=-3\left(x-\left(-4\right)\right)\\ y-2=-3\left(x+4\right)\\ y=-3x-12+2\\ y=-3x-10\end{array}$

Thus, the equation of the tangent line is $y=-3x-10$ .

c.

To determine

Tograph : The function $f\left(x\right)=x^2+5x+6$ and its tangent line.

Explanation of Solution

Given information : The given function is $f\left(x\right)=x^2+5x+6$ .

Graph :

As derived in part b, the equation of the tangent line is $y=-3x-10$ .

  

Interpretation : The graph of the function and the tangent line intersects at a point, $\left(-4,2\right)$ .