Question

a. determine the points where the curve 5x + 5y² - y = 2 has a vertical tangent line.
b. does the curve have any horizontal tangent lines?
a. the curve has a vertical tangent line at the point(s) . (type an ordered pair. use a comma to separate answers as needed. use integers or fractions for any numbers in the expression )

Explanation:

Step1: Differentiate implicitly

Differentiate $5x + 5y^{2}-y = 2$ with respect to $x$.
$5+10y\frac{dy}{dx}-\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(10y - 1)=- 5$, so $\frac{dy}{dx}=\frac{-5}{10y - 1}$.

Step3: Find vertical - tangent points

A vertical tangent occurs when $\frac{dy}{dx}$ is undefined, i.e., when the denominator is $0$.
Set $10y - 1=0$, then $y=\frac{1}{10}$.
Substitute $y = \frac{1}{10}$ into the original equation $5x+5y^{2}-y = 2$.
$5x+5(\frac{1}{10})^{2}-\frac{1}{10}=2$.
$5x+\frac{5}{100}-\frac{1}{10}=2$.
$5x+\frac{1}{20}-\frac{2}{20}=2$.
$5x-\frac{1}{20}=2$.
$5x=2+\frac{1}{20}=\frac{40 + 1}{20}=\frac{41}{20}$.
$x=\frac{41}{100}$.
The point is $(\frac{41}{100},\frac{1}{10})$.

Step4: Find horizontal - tangent points

A horizontal tangent occurs when $\frac{dy}{dx}=0$. Since the numerator of $\frac{dy}{dx}=\frac{-5}{10y - 1}$ is non - zero ($- 5
eq0$), there are no horizontal tangent lines.

Answer:

a. $(\frac{41}{100},\frac{1}{10})$
b. No