Question

Determine the points where the curve 2x+4y^2-y=1has a vertical tangent line

Answer 1

The given equation of curve is

$$2x+4y^2-y=1$$

Since, vertical tangent found to those points where slope=$dy/dx$ is $\pm \infty$

Now, take derivative both sides of the above curve with respect to x we get

$$2+8y\frac{dy}{dx}-\frac{dy}{dx}=0\\ \implies \frac{dy}{dx}=\frac{2}{1-8y}$$

Thus,$\frac{dy}{dx}=\pm\infty$ only at $y=1/8$ , hence on plugin this value to the equation of curve we get,

$$2x+4(1/8)^2-(1/8)=1\\ \implies 2x=\frac{17}{16}\implies x=\frac{17}{32}$$

Thus, point where the curve $2x+4y^2-y=1$ has vertical tangent is $(17/32,1/8)$