$y(x)$ at $x = 2$. But, as you learned in algebra, the slope of $y_{t}(x)$ is $b$. The slope of $y(x)$ at any point $x$ is $\frac{dy(x)}{dx}$\
(3) At $x=2$, $\frac{dy(x)}{dx}= \frac{5}{7}x^{4} =
\frac{80}{7}$. So $b = \frac{80}{7}$. (4) Also, $y_{t}(x)$ and $y(x)$ must be equal at the point of tangency, since they share the same point. So $ a + \frac{80}{7}x = \frac{1}{7}x^5$ at $x
= 2$, or $a + \frac{160}{7} = \frac{32}{7}$. Thus, $a = - \frac{128}{7}$. (5) The solution is $y_{t}(x) = \frac{-128 + 80x}{7}$. Fig. 2 is a graph of the solution showing the tangent line and the curve.

Recall why we are interested in the slopes of tangents to curves. We began our discussion of calculus by demonstrating that “the slope of a line on a distance-time graph …corresponds to velocity". So if we are given a function that relates distance to time, the derivative of that function, being the the slope of the function, corresponds to the velocity as a function of time. The next exercise makes use of that relationship.

[**EXERCISE**]{}

[**2.8**]{} (a) A rock is hurled upward from a platform that is $H$ meters ($Hm$) above the ground with an initial velocity 