Curve with $\frac{dy}{dx}=-\frac{15}{(5x-2)^2}$ through $(4,-2)$

Key takeaway: This is a standard calculus question that combines integration of a derivative function with graph transformations. First find the original curve from the given gradient and point, then apply the stretch and translation carefully in the correct order.

(a) Find the equation of the curve

Given

$\frac{dy}{dx}=-\frac{15}{(5x-2)^2}$

Integrate with respect to $x$:

$y=\int -\frac{15}{(5x-2)^2}\,dx$

Let $u=5x-2$, so $du=5\,dx$ and $dx=\frac{du}{5}$.

Then

$y=\int -15u^{-2}\cdot \frac{du}{5}=-3\int u^{-2}\,du$

$y=-3\left(\frac{u^{-1}}{-1}\right)+C=\frac{3}{u}+C$

So

$y=\frac{3}{5x-2}+C$

Use the point $(4,-2)$:

$-2=\frac{3}{5(4)-2}+C$

$-2=\frac{3}{18}+C=\frac{1}{6}+C$

$C=-\frac{13}{6}$

Hence the equation of the curve is

$\boxed{y=\frac{3}{5x-2}-\frac{13}{6}}$

(b) Stretch in the $x$-direction by scale factor 2, then translate by $\left(-\frac17\right)$

A stretch in the $x$-direction by factor 2 means replace $x$ by $\frac{x}{2}$:

$y=\frac{3}{5\left(\frac{x}{2}\right)-2}-\frac{13}{6}$

$y=\frac{3}{\frac{5x}{2}-2}-\frac{13}{6}$

Now simplify the denominator:

$\frac{5x}{2}-2=\frac{5x-4}{2}$

So

$y=\frac{3}{(5x-4)/2}-\frac{13}{6}=\frac{6}{5x-4}-\frac{13}{6}$

A translation of $-\frac17$ in the $y$-direction means move down by $\frac17$:

$y=\frac{6}{5x-4}-\frac{13}{6}-\frac17$

Combine the constants:

$-\frac{13}{6}-\frac17=-\frac{91}{42}-\frac{6}{42}=-\frac{97}{42}$

So the new curve is

$\boxed{y=\frac{6}{5x-4}-\frac{97}{42}}$

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Pitfalls the pros know 👇 A common error is to integrate the derivative and forget the constant of integration. Another frequent mistake is applying the horizontal stretch incorrectly: a stretch in the $x$-direction by factor 2 replaces $x$ with $x/2$, not $2x$. For the final translation, make sure the vertical shift is applied after the stretch.

What if the problem changes? If the stretch factor in the $x$-direction were $k$, the transformed equation would come from replacing $x$ by $x/k$ first, then applying any vertical translation. If the translation were a vector $\begin{pmatrix}p\\q\end{pmatrix}$, the final step would be $y=f(x-p)+q$ after the horizontal transformation.

Tags: integration, gradient function, graph transformation