Find the slope of the curve $y=x^2-4x-5$ 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 $(3+h, x^2-4(3+h)^2-5)$.
The secant slope is $$\frac{\Delta y}{\Delta x}$$ $$\implies \frac{[x^2-4(3+h)^2-5]- [x^2-4(3)-5]}{(3+h)-3}$$
$$\frac{[x^2-4(h^2+6h+9)-5] – [x^2-17]}{h}$$
$$\frac{[x^2-4h^2-24h-36-5] – [x^2-17]}{h}$$
$$\frac{-4h^2-24h-24}{h}$$
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.
Best Answer
Your slope calculation is off a bit, you should have
$${[(3+h)^2-4(3+h)-5]-[3^2-4(3)-5]\over (3+h)-3}$$
$$={2h+h^2\over h}$$
This is because $Q$ is located at $(3+h,(3+h)^2-4(3+h)-5)$.
Then yes, take the limit as $h\to 0$ and you have the slope.