Chapter 3.1, Problem 18E

To determine

Tocalculate:The derivatives of the function $y=2x^2-13x+5$ and the equation of the line tangent to the curve at $x=3$ .

Answer to Problem 18E

The required derivative of $y=2x^2-13x+5$ is $4x-13$ and the equation of the tangent line is $y=-x-13$ .

Explanation of Solution

Given information:

The algebraic equation $y=2x^2-13x+5$ .

Point on the tangent $x=3$ .

Formula used:

First order differentiation rule:

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

Where $f\left(x\right)$ is the function and $h$ is constant term.

General form of a straight line:

  $\left(y-y_1\right)=m\left(x-x_1\right)$

Calculation:

The given algebraic equation is $y=2x^2-13x+5$ .

Find the derivative of the above equation.

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{[2{\left(x+h\right)}^2--13\left(x+h\right)+5]-\left(2x^2-13x+5\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{4xh+2h^2-13h}{h}\\ =\underset{h\to 0}{\lim }\left(4x+2h-13\right)\\ =4x-13\end{array}$

Substitute the value 3 for $x$ .

  $\begin{array}{c}\frac{dy}{dx}=4x-13\\ =4\times 3-13\\ =12-13\\ =-1\end{array}$

Thus, the slop of the tangent is $-1$ .

Calculate the y intercept.

Substitute the value 3 for $x$ in the equation $y=2x^2-13x+5$ .

  $\begin{array}{c}y=2\times 3^2-13\times 3+5\\ =2\times 9-39+5\\ =18-39+5\\ =-16\end{array}$

It is known that the general form of a straight line is $\left(y-y_1\right)=m\left(x-x_1\right)$ .

Substitute the value $3,-16\text{and}-1$ for $x,y\text{and}m$ respectively.

  $\begin{array}{c}\left(y-\left(-16\right)\right)=-1\left(x-3\right)\\ \left(y+16\right)=-\left(x-3\right)\\ y=-x-13\end{array}$

Thus, the equation of the tangent line is $y=-x-13$ .

Hence, the required derivative of $y=2x^2-13x+5$ is $4x-13$ and the equation of the tangent line is $y=-x-13$ .