The curve for which $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{a}{(2x+3)^3} - 1$ where $a$ is a co

Question

The curve for which $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{a}{(2x+3)^3} - 1$ where $a$ is a constant, is such that the tangent to the curve at $(-1, 0)$ is perpendicular to the line $5y = x + 1$. Find the value of $a$ and the equation of the curve.

Accepted Answer

$a=-4$, $y=\frac{1}{(2x+3)^2}-x-2$

Differentiation - Tangents and Normals at a Point

Differentiation - Tangents and Normals at a Point

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