I'm supposed to get the equation of the tangent line to the graph of $f(x)= \frac{8}{x}$ at the point $(2,4)$.
I started with
$$\frac{\frac{8}{x+h} – \frac{8}{x}}{h},$$
then I cross multiplied:
$$\frac{8x – 8(x+h)}{
x(x+h)}$$
Where do I go from here?
Best Answer
Actually you forgot the $h$ term in the denominator. We have $\frac{8x-8(x+h)}{x(x+h)h}=\frac{-8h}{x(x+h)h}$ and this simplifies to $\frac{-8}{x(x+h)}$. We need to take the limit of this term as $h \rightarrow 0$. This is simply $\frac{-8}{x^2}$, which is exactly the derivative of the curve.
Evaluate the expression $\frac{-8}{x^2}$ at $x=2$ for the exact numerical value of the slope of the tangent at the given point. Then apply the fact that a straight line is uniquely determined by its slope and a point it passes through using the equation $(y-y_0)=m \times (x-x_0)$, where $m$ is the slope, which we computed earlier, and $(x_0, y_0)$ is the point you already mentioned in the question. Does that make sense?