$f(x) = 8 b^x$
My answer will be in terms of $b$.
I know you find the slope typically with $\dfrac{y_1-y_2}{x_1-x_2}$, but that doesn't seem to work in this situation.
$f(x) = 8 b^x$
My answer will be in terms of $b$.
I know you find the slope typically with $\dfrac{y_1-y_2}{x_1-x_2}$, but that doesn't seem to work in this situation.
Write $\,Q=(5,8\cdot b^5)\,\,,\,P=(1,8\cdot b)\,$ , then the slope between these two points is:
$m_{PQ}=\frac{8b^5-8b}{5-1}=2b(b^4-1)$
It looks like the original question had no calculus tag and judging by the way the question was asked, I'm guessing no knowledge of calculus.
The formula you posted for slope works for a line where slope is constant. Without knowledge of calculus, the best you can do is try to draw an approximate tangent line, the find the slope of the line.
This is a problem made for calculus though. The slope at any point on a curve $y=f(x)$ is the derivative of $y$ with respect to $x$, written as $\frac{dy}{dx}$. Conan Wong gives the correct calculation for this derivative.
$b^x = e^{ln (b^x)} = e^{x ln (b)}$
so using the Chain Rule and the fact that $(e^x)' = e^x$ we have
$(b^x)' = (e^{xln(b)})' = (ln(b)) e^{xln(b)} = (ln (b))b^x$
So the derivative of $8b^x$ is $8(ln(b))b^x$.