Slope of a curve

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Two young mathematicians discuss the novel idea of the “slope of a curve.”

Check out this dialogue between two calculus students (based on a true story):

Devyn: Riley, do you remember “slope?’
Riley: Most definitely. “Rise over run.”
Devyn: You know it.
Riley: “Change in $y$ over change in $x$ .”
Devyn: That’s right.
Riley: Brought to you by the letter “ $m$ .”
Devyn: Enough! My important question is: could we define “slope” for a curve that’s not a straight line?
Riley: Well, maybe if we “zoom in” on a curve, it would look like a line, and then we could call it “the slope at that point.”
Devyn: Ah! And this “zoom in” idea sounds like a limit!
Riley: This is so awesome. We just made math!

The concept introduced above, of the “slope of a curve at a point,” is in fact one of the central concepts of calculus. It will, of course, be completely explained. Let’s explore Devyn and Riley’s ideas a little more, first.

To find the “slope of a curve at a point,” Devyn and Riley spoke of “zooming in” on a curve until it looks like a line. When you zoom in on a smooth curve, it will eventually look like a line. This line is called the tangent line.

Which of the following approximate the slope of the “zoomed line”? (You can select more than one.)

$\frac {f(x_1)-f(a)}{x_2-a}$ $\frac {f(x_1)-f(a)}{a-x_1}$ $\frac {f(x_1)-f(a)}{x_1-a}$ $\frac {f(x_1)-a}{x_1-a}$ $\frac {f(a)-f(x_2)}{a-x_2}$ $\frac {f(x_2)-f(a)}{x_2-a}$ $\frac {x_2-a}{f(x_2)-f(a)}$ $\frac {f(x_1)-f(a)}{f(x_2)-f(a)}$ $a(f(x_2)-f(x_1))$ $\frac {f(a)-f(x_1)}{a-x_1}$

Let $f(x) = 3x-1$ . Zoom in on the curve around $a = -2$ so that $x_1 = -1.9$ . Use one of the formulations in the problem above to approximate the slope of the curve. The slope of the curve at $a = -2$ is approximately… $\answer {3}$

Repeat the previous problem for $f(x) = x^2 - 1$ , $a = 0$ , and $x_2 = 0.2$ . Choose a formulation that will give you a positive answer for the slope. The (positive) slope of the curve at $a = 0$ is approximately… $\answer {0.2}$

Zoom in on the curve $f(x) = x^2 - 1$ near $x=0$ again. By looking at the graph, what is your best guess for the actual slope of the curve at zero?

impossible to say zero one infinity