Question

a. determine the points where the curve 6x + 2y^2 - y = 3 has a vertical tangent line.
(type an ordered pair. use a comma to separate answers as needed. use integers or fractions for any numbers in the expression.)

b. does the curve have any horizontal tangent lines?

Explanation:

Step1: Differentiate the given equation implicitly

Differentiate $6x + 2y^{2}-y=3$ with respect to $x$.
The derivative of $6x$ with respect to $x$ is $6$, the derivative of $2y^{2}$ using the chain - rule is $4y\frac{dy}{dx}$, and the derivative of $-y$ is $-\frac{dy}{dx}$, and the derivative of the constant $3$ is $0$. So, $6 + 4y\frac{dy}{dx}-\frac{dy}{dx}=0$.

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

Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(4y - 1)=-6$. Then $\frac{dy}{dx}=\frac{-6}{4y - 1}$.

Step3: Find horizontal tangent points

A horizontal tangent line occurs when $\frac{dy}{dx}=0$. Since the numerator of $\frac{dy}{dx}=\frac{-6}{4y - 1}$ is non - zero ($-6
eq0$), there are no horizontal tangent lines.

Step4: Find vertical tangent points

A vertical tangent line occurs when the denominator of $\frac{dy}{dx}$ is zero. Set $4y - 1 = 0$. Solving for $y$, we get $y=\frac{1}{4}$.
Substitute $y = \frac{1}{4}$ into the original equation $6x+2y^{2}-y = 3$.
$6x+2(\frac{1}{4})^{2}-\frac{1}{4}=3$.
$6x+2\times\frac{1}{16}-\frac{1}{4}=3$.
$6x+\frac{1}{8}-\frac{1}{4}=3$.
$6x-\frac{1}{8}=3$.
$6x=3+\frac{1}{8}=\frac{24 + 1}{8}=\frac{25}{8}$.
$x=\frac{25}{48}$.

Answer:

a. $(\frac{25}{48},\frac{1}{4})$
b. No