Curve sketching uses calculus to turn an equation into a reliable picture of its graph. Instead of plotting many random points, you identify the structure of the function through domain, intercepts, asymptotes, derivatives, and concavity. This matters because the graph shows where a quantity increases, decreases, levels off, blows up, or changes shape.
A good sketch also helps you check algebraic work and understand the behavior of functions that are hard to visualize from the formula alone.
For the worked example f(x) = (x^2 - 1)/(x^2 - 4), the graph has holes in the domain at x = -2 and x = 2, which create vertical asymptotes. Rewriting the function as f(x) = 1 + 3/(x^2 - 4) reveals the horizontal asymptote y = 1 and helps explain the end behavior. The first derivative f'(x) = -6x/(x^2 - 4)^2 shows increasing and decreasing intervals, while the second derivative f''(x) = 6(3x^2 + 4)/(x^2 - 4)^3 shows concavity.
Together, these features produce a complete, labeled sketch with intercepts, asymptotes, a local maximum, and separate curve branches.
Key Facts
- Domain of f(x) = (x^2 - 1)/(x^2 - 4) is x != -2 and x != 2.
- x-intercepts occur where x^2 - 1 = 0, so x = -1 and x = 1.
- The y-intercept is f(0) = 1/4.
- Vertical asymptotes occur at x = -2 and x = 2.
- Since f(x) = 1 + 3/(x^2 - 4), the horizontal asymptote is y = 1.
- f'(x) = -6x/(x^2 - 4)^2 and f''(x) = 6(3x^2 + 4)/(x^2 - 4)^3.
Vocabulary
- Domain
- The domain is the set of all input values x for which a function is defined.
- Vertical asymptote
- A vertical asymptote is a vertical line that the graph approaches as the function grows without bound or decreases without bound.
- Horizontal asymptote
- A horizontal asymptote is a horizontal line that the graph approaches as x goes toward positive or negative infinity.
- Critical point
- A critical point is a point in the domain where f'(x) = 0 or where f'(x) does not exist.
- Concavity
- Concavity describes whether a graph bends upward like a cup or downward like a cap.
Common Mistakes to Avoid
- Canceling x^2 - 1 with x^2 - 4, which is wrong because these are different expressions and have no common factor.
- Forgetting to exclude x = -2 and x = 2 from the domain, which is wrong because the denominator is zero at those values.
- Calling x = -2 and x = 2 critical points, which is wrong because critical points must be in the domain of the function.
- Marking inflection points at the vertical asymptotes, which is wrong because inflection points must be actual points on the graph where concavity changes.
Practice Questions
- 1 For f(x) = (x^2 - 1)/(x^2 - 4), find the domain, x-intercepts, and y-intercept.
- 2 For f(x) = (x^2 - 1)/(x^2 - 4), use f'(x) = -6x/(x^2 - 4)^2 to find the intervals where the function is increasing and decreasing.
- 3 Explain why x = -2 and x = 2 are vertical asymptotes but not inflection points for f(x) = (x^2 - 1)/(x^2 - 4).