Question

y = 2x² - ln(x), (1, 2)
(a) find an equation of the tangent line to the graph of the function at the given point.
y =
(b) use a graphing utility to graph the function and its tangent line at the point. use the tangent feature of a graphing utility to verify your results. (select update graph to see your response plotted on the screen. select the submit button to grade your response.)

Explanation:

Step1: Find derivative of the function

The derivative of $y=2x^2-\ln(x)$ is $y' = 4x - \frac{1}{x}$

Step2: Calculate slope at $x=1$

Substitute $x=1$ into $y'$:
$y'(1) = 4(1) - \frac{1}{1} = 4 - 1 = 3$

Step3: Use point-slope form

Point-slope formula: $y - y_1 = m(x - x_1)$, where $(x_1,y_1)=(1,2)$ and $m=3$
$y - 2 = 3(x - 1)$

Step4: Simplify to slope-intercept form

$y - 2 = 3x - 3$
$y = 3x - 1$

Answer:

$y = 3x - 1$

(Note: Part (b) requires a graphing utility to complete, which cannot be done in this text format, but the equation for the tangent line to graph is provided above)