# Curve sketching

Imagine you are taking a Calculus exam. The only question remaining is to show what the graph of f(x) looks like. Just as your turn your graphing calculator on, all electricity in the room goes out including all calculators and cell phones. Now what?

For the following tasks, graphing calculators will not be allowed.

Part I. Complete the table below using the following words: concave down, concave up, critical number, decreasing, increasing, possible inflection point

 Given $$f(x)$$ is … $$f'(x) < 0$$ $$f'(x) = 0$$ $$f'(x) = DNE$$ $$f'(x) > 0$$ $$f''(x) < 0$$ $$f''(x) = 0$$ $$f''(x) = DNE$$ $$f''(x) > 0$$

Part II. Draw the general shape of a graph given the following information.

 a. $$f' \gt 0 \text{ and } f'' \gt 0$$ b. $$f' \gt 0 \text{ and } f'' \lt 0$$ c. $$f' \lt 0 \text{ and } f'' \gt 0$$ d. $$f' \lt 0 \text{ and } f'' \lt 0$$

Part III. Consider the following function: $$f(x) = \frac{x^2}{(x-2)^2}$$ and its graph $$y=f(x)$$. Find the following:

1. Domain
2. Vertical asymptotes
3. Horizontal asymptotes
4. Critical points
5. Intervals where increasing/decreasing
6. Possible inflection points
7. Intervals where concave up/down
8. Inflection points
9. Intercepts

Sketch the graph based on the above information. When done, go ahead and see how you did on your graphing calculator.

1. $$x \neq 2$$
2. $$\displaystyle \lim_{x \to 2^-} f(x) = \infty, \displaystyle \lim_{x \to 2^+} f(x) = \infty$$, so vertical asymptote at $$x = 2$$
3. $$\displaystyle \lim_{x \to \infty} f(x) = 1, \displaystyle \lim_{x \to -\infty} f(x) = 1$$, so horizontal asymptote at $$y = 1$$
4. $$f'(x) = \frac{-4x}{(x-2)^3}$$ with critical points at at $$x = 0, 2$$
5. decreasing: $$x \lt 0 \text{ or } x \gt 2$$
increasing: $$0 \lt x \lt 2$$
6. $$f''(x) = \frac{8x+8}{(x-2)^4}$$ with possible inceptions points at $$x = -1, 2$$
7. concave down: $$x \lt -1$$
concave up: $$-1 \lt x \lt 2$$ or $$x \gt 2$$
8. at $$x=-1$$ concavity changes due to $$f''$$ changing signs from negative to positive
9. $$x=0$$ implies $$(0,0)$$
$$y=0$$ implies $$(0,0)$$