# 007A Sample Midterm 1, Problem 3 Detailed Solution

Let  ${\displaystyle y=2x^{2}-3x+1.}$

(a) Use the definition of the derivative to compute   ${\displaystyle {\frac {dy}{dx}}.}$

(b) Find the equation of the tangent line to  ${\displaystyle y=2x^{2}-3x+1}$  at  ${\displaystyle (2,3).}$

Background Information:
Recall
${\displaystyle f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$

Solution:

(a)

Step 1:
Let  ${\displaystyle f(x)=2x^{2}-3x+1.}$
Using the limit definition of the derivative, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {2(x+h)^{2}-3(x+h)+1-(2x^{2}-3x+1)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {2x^{2}+4xh+2h^{2}-3x-3h+1-2x^{2}+3x-1}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {4xh+2h^{2}-3h}{h}}.}\end{array}}}$

Step 2:
Now, we simplify to get
${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {h(4x+2h-3)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}(4x+2h-3)}\\&&\\&=&\displaystyle {4x-3.}\end{array}}}$

(b)

Step 1:
We start by finding the slope of the tangent line to  ${\displaystyle f(x)=2x^{2}-3x+1}$  at  ${\displaystyle (2,3).}$
Using the derivative calculated in part (a), the slope is
${\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {f'(2)}\\&&\\&=&\displaystyle {4(2)-3}\\&&\\&=&\displaystyle {5.}\end{array}}}$
Step 2:
Now, the tangent line to  ${\displaystyle f(x)=2x^{2}-3x+1}$  at  ${\displaystyle (2,3)}$
has slope  ${\displaystyle m=5}$  and passes through the point  ${\displaystyle (2,3).}$
Hence, the equation of this line is
${\displaystyle y=5(x-2)+3.}$
If we simplify, we get
${\displaystyle y=5x-7.}$

(a)     ${\displaystyle {\frac {dy}{dx}}=4x-3}$
(b)     ${\displaystyle y=5x-7}$