Grade 12 · MCV4U · Calculus & Vectors

MCV4U Curve Sketching Practice Test (with Solutions)

Q: How do you find the absolute max and min on a closed interval?

Find the critical points inside the interval, then evaluate f at each of them and at both endpoints. The largest value is the absolute maximum and the smallest is the absolute minimum.

Curve Sketching is where every calculus tool so far comes together on one graph. The free test takes a single rational function, f(x) = (x - 4)/(x² - x - 2), and walks it through a full sketch: domain, intercepts, asymptotes analysed with limits, critical points and points of inflection. Do each step below on paper, then check it against the solution.

Free Curve Sketching Practice Test

8 questions, exactly as a free account gets them in the first Curve Sketching test on Martian Lab. Try each one on paper first, then open the answer.

Given $f(x) = \dfrac{x-4}{x^2-x-2}$ , complete the following over its entire domain. Domain: $\rule{1cm}{0.4pt}$
Show answer
$D : \{ x \in \mathbb{R},\ x \neq 2,\ -1 \}$
Given $f(x) = \dfrac{x-4}{x^2-x-2}$ , complete the following over its entire domain. y-intercept: $\rule{1cm}{0.4pt}$
Show answer
$x = 0,\ y = 2$ $y\text{-intercept at }(0,\ 2)$
Given $f(x) = \dfrac{x-4}{x^2-x-2}$ , complete the following over its entire domain. x-intercept(s): $\rule{1cm}{0.4pt}$
Show answer
$y = 0\ ,\ x = 4$ $x\text{-intercept at }(4,\ 0)$
Given $f(x) = \dfrac{x-4}{x^2-x-2}$ , complete the following over its entire domain. State the equation of the vertical asymptote(s) and use limits to analyze the behaviours of $f(x)$ near its vertical asymptote.
Show solution
$V.A. :\ x = 2\ ,\ x = -1$ $\displaystyle\lim_{x \to 2^+} f(x) = -\infty$ $\displaystyle\lim_{x \to 2^-} f(x) = +\infty$ $\displaystyle\lim_{x \to -1^+} f(x) = +\infty$ $\displaystyle\lim_{x \to -1^-} f(x) = -\infty$
Given $f(x) = \dfrac{x-4}{x^2-x-2}$ , complete the following over its entire domain. State the equation of the horizontal asymptote and use limits to analyze the end behaviours of $f(x)$
Show solution
$\displaystyle\lim_{x \to \infty} \dfrac{x-4}{x^2-x-2}$ $\displaystyle= \lim_{x \to \infty} \dfrac{x^2\!\left(\dfrac{1}{x} - \dfrac{4}{x^2}\right)}{x^2\!\left(1 - \dfrac{1}{x} - \dfrac{2}{x^2}\right)}$ $\displaystyle= \lim_{x \to \infty} \dfrac{0}{1}$ $= 0$ $\therefore\ H.A\ \text{is}\ y = 0$ $\lim_{x \to \infty} f(x) = 0^+$ $\therefore\ \text{as}\ x \to \infty,\ y \to 0\ \text{from above}$ $\lim_{x \to -\infty} f(x) = 0^-$ $\therefore\ \text{as}\ x \to -\infty,\ y \to 0\ \text{from below}$
Given $f(x) = \dfrac{x-4}{x^2-x-2}$ , complete the following over its entire domain. Determine if there is any cross over point, if so, locate it.
Show answer
Cross over: point of intersection between function and H.A. $\dfrac{x-4}{x^2-x-2} = 0\ ,\ x=4$ $\therefore$ Cross over point at $(4,\ 0)$ .
Given $f(x) = \dfrac{x-4}{x^2-x-2}$ , complete the following over its entire domain. Locate all critical points. You may round to 1 decimal places.
Show solution
$f'(x) = \dfrac{1\cdot(x^2-x-2)-(x-4)(2x-1)}{(x^2-x-2)^2}$ $= \dfrac{x^2-x-2-2x^2+9x-4}{(x^2-x-2)^2}$ $= \dfrac{-x^2+8x-6}{(x^2-x-2)^2}$ $\therefore\ f'(x)=0$ $x = \dfrac{-8 \pm \sqrt{40}}{-2}$ $= 4 \pm \sqrt{10}$ $x \approx 7.16 \quad \text{or} \quad 0.84$
Given $f(x) = \dfrac{x-4}{x^2-x-2}$ , complete the following over its entire domain. Locate all P.O.I. by showing that $f''(10.398) \approx 0$
Show solution
$f''(x) = \dfrac{(-2x+8)(x^2-x-2)^2 - (-x^2+8x-6)2(x^2-x-2)(2x-1)}{(x^2-x-2)^4}$ $= \dfrac{2(x^2-8x+6)(2x-1) - 2(x-4)(x^2-x-2)}{(x^2-x-2)^3}$ $= \dfrac{2\left[2x^3-16x^2+12x-x^2+8x-6-x^3+4x^2+x^2-4x+2x-8\right]}{(x^2-x-2)^3}$ $= \dfrac{2(x^3-12x^2+18x-14)}{(x^2-x-2)^3}$ $\therefore\ f''(10.398) \approx 0$

This Unit on Martian Lab

The lessons go from maximums and minimums, with and without boundaries, through concavity and the second derivative test to full curve sketches and graph analysis.

Maximums and Minimums
Maximums and Minimumswith Boundaries
Concavity and Second Derivative Test
Curve Sketching
Graph Analysis

Then 4 full Curve Sketching practice tests to check yourself.

Start Curve Sketching free

Curve Sketching FAQ

What is on an MCV4U curve sketching test?+

Local and absolute maximums and minimums, intervals of increase and decrease, concavity and points of inflection, the second derivative test, and full sketches of polynomial and rational functions using limits at the asymptotes. The free test on this page runs one rational function through the whole process.

What are the steps to sketch a curve?+

Domain, then intercepts, then asymptotes: vertical ones with one-sided limits and horizontal ones with limits at infinity. Use f' to find critical points and where f increases or decreases, and f'' to find concavity and points of inflection. Plot all of it, then join the pieces.

What is a critical point?+

A point in the domain where f'(x) = 0 or f'(x) does not exist. It is only a candidate: check the sign of f' on either side, or use the second derivative test, to decide whether it is a maximum, a minimum or neither.

How does the second derivative test work?+

At a critical point where f'(c) = 0, if f''(c) > 0 the graph is concave up and c is a local minimum. If f''(c) < 0 it is concave down and c is a local maximum. If f''(c) = 0 the test tells you nothing, so go back to the first derivative test.

How do you find the absolute max and min on a closed interval?+

Find the critical points inside the interval, then evaluate f at each of them and at both endpoints. The largest value is the absolute maximum and the smallest is the absolute minimum.

Curve Sketching Topics

One idea at a time: a short explanation, then five worked examples.

Curve Sketching Steps

More MCV4U Units

Previous unit: Derivatives

Next unit: Optimization

Every MCV4U unit and a free sample test

Start Practicing

No credit card required to start.

Get started