Chapter 3.1, Problem 1E

(a)

To determine

The slope of the tangent line to the graph of ‘ $f$ ’ at $P\left(a,f\left(a\right)\right)$

Answer to Problem 1E

  $m_a=10a-4$

Explanation of Solution

Given:

The given expression is

  $f\left(x\right)=5x^2-4x$

Calculation:

The slope of the tangent line to the graph can be calculated as

  $\begin{array}{l}{m}_a=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ m_a=\underset{h\to 0}{\lim }\frac{\left(5{\left(a+h\right)}^2-4\left(a+h\right)\right)-\left(5a^2-4a\right)}{h}\\ m_a=\underset{h\to 0}{\lim }\frac{\left(5a^2+10ah-4a+5h^2-4h\right)-\left(5a^2-4a\right)}{h}\\ m_a=\underset{h\to 0}{\lim }\frac{10ah+5h^2-4h}{h}\\ m_a=\underset{h\to 0}{\lim }\frac{h\left(10a+5h-4\right)}{h}\\ m_a=\underset{h\to 0}{\lim }\left(10a+5h-4\right)\\ m_a=10a+5\left(0\right)-4\\ m_a=10a-4\end{array}$

(b)

To determine

The equation of the tangent line at $P\left(2,f\left(2\right)\right)$

Answer to Problem 1E

  $y-16x+20=0$

Explanation of Solution

Given:

The given expression is

  $f\left(x\right)=5x^2-4x$

Calculation:

The equation of the tangent line can be calculated as

  $\begin{array}{l}f\left(x\right)=5x^2-4x\\ f\left(2\right)=5{\left(2\right)}^2-4\left(2\right)\\ f\left(2\right)=20-8\\ f\left(2\right)=12\end{array}$

Therefore,

  $P\left(2,f\left(2\right)\right)=P\left(2,12\right)$

Slope of the equation of the tangent line is $m_a=10a-4$

  $\begin{array}{l}{m}_2=10\left(2\right)-4\\ m_2=16\end{array}$

Equation of the tangent at $P\left(2,12\right)$ will be

  $\begin{array}{l}y-y_1=m_a\left(x-x_1\right)\\ y-12=16\left(x-2\right)\\ y-12=16x-32\\ y-16x=12-32\\ y-16x=-20\\ y-16x+20=0\end{array}$