Finding slope of a curve by finding the limits of secant slopes
Your slope calculation is off a bit, you should have
$${[(3+h)^24(3+h)5][3^24(3)5]\over (3+h)3}$$
$$={2h+h^2\over h}$$
This is because $Q$ is located at $(3+h,(3+h)^24(3+h)5)$.
Then yes, take the limit as $h\to 0$ and you have the slope.
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Emi Matro
Updated on January 29, 2020Comments

Emi Matro almost 4 years
Find the slope of the curve $y=x^24x5$ at the point $P(3,8)$ by finding the limit of the secant slopes through point $P$.
My try:
I picked another point $Q$ to get the secant $PQ$. Since $P$ is $(3,8)$, $Q$ is $(3+h, x^24(3+h)^25)$.
The secant slope is $$\frac{\Delta y}{\Delta x}$$ $$\implies \frac{[x^24(3+h)^25] [x^24(3)5]}{(3+h)3}$$
$$\frac{[x^24(h^2+6h+9)5]  [x^217]}{h}$$
$$\frac{[x^24h^224h365]  [x^217]}{h}$$
$$\frac{4h^224h24}{h}$$
and this is where I'm lost. The answer is $2$, and I don't see how they got that. The limit of the secant slopes through $P$ means that $h$ is getting closer to $0$, right? Please help and explain. Thanks.