Finding the equation of a Tangent line to a graph at a specific point

Question

Trying to determine the tangent line to the graph of F(x)=x^3 at (2,8). I know you use the equation (F(Dx+x)-f(x))/Dx for this type of problem, and that you just plug in the x-value at the point to solve once you have the simplified equation; I am having trouble simplifying to remove the Delta-x values, though.

Answer 1

By definition, the tangent line of a function at a point is its slope, or more specifically, its derivative. Using limits to find a derivative, we have $$\frac{f(x+\Delta) - f(x)}{\Delta}.$$

When $f(x) = x^3$, from the binomial expansion $f(x+\Delta) = (x+\Delta )^3 = x^3+x^2\Delta+x \Delta^2+ \Delta^3$. Now we can substitute those values in, and we have:

$$\frac{x^3+3x^2\Delta+3x \Delta^2+ \Delta^3 - x^3}{\Delta}$$ $$=\frac{3x^2\Delta+3x \Delta^2+ \Delta^3}{\Delta}$$ $$=3x^2+3x\Delta+\Delta^2.$$

Since as $\Delta$ tends to $0$, any term with $\Delta$ will also tend to $0$, the only term without a $\Delta$ is $3x^2$, which is the derivative of $x^3$.

Now find the point's $x$-coordinate and plug that into $3x^2$ to get the slope at that point.