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Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Full Curve Sketching

A full curve sketch means drawing a function from its equation alone, with every feature that matters in the right place. Each feature comes from one test, and a sign chart puts them together.

How It Works

Work through the same checklist every time. The order matters, because each step hands the next one the numbers it needs.

1. Domain. Note anything excluded, like a zero in a denominator or a negative under an even root.

2. Intercepts. Set x=0x = 0 for the yy-intercept. Set y=0y = 0 and factor for the xx-intercepts.

3. Asymptotes. A vertical asymptote sits where the denominator is zero and the numerator is not. If both are zero, the factor cancels and you get a hole instead. Compare degrees for end behaviour: a smaller numerator gives y=0y = 0, equal degrees give the ratio of leading coefficients, and a numerator one degree higher gives an oblique asymptote that you find by long division.

4. Symmetry. If f(−x)=f(x)f(-x) = f(x) the graph is even; if f(−x)=−f(x)f(-x) = -f(x) it is odd. Either one halves the work.

5. First derivative. Find where f′f' is zero or undefined. These critical numbers split the line into intervals. Test a point in each: positive means increasing, negative means decreasing. A sign change from plus to minus is a local max, minus to plus is a local min, and no change means neither.

6. Second derivative. Find where f′′f'' is zero or undefined. Positive means concave up, negative means concave down. A change in concavity at a point on the curve is a point of inflection.

7. Sign chart, then sketch. Put every special xx-value on one chart with rows for f′f' and f′′f''. Plot the points you found, draw the asymptotes as dashed lines, then join the pieces with the right direction and bend.

A derivative that is undefined at a point on the curve is a cusp or a vertical tangent, so give it a sharp point or a vertical edge.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

  1. Easy

    Sketch y=x3−3xy = x^3 - 3x. Find the intercepts, local extrema and point of inflection.
    Show solution
    1. Intercepts: x(x2−3)=0x(x^2 - 3) = 0 gives x=0,±3x = 0, \pm\sqrt{3}; the yy-intercept is 00. The function is odd.
    2. y′=3x2−3=3(x−1)(x+1)y' = 3x^2 - 3 = 3(x - 1)(x + 1), zero at x=±1x = \pm 1. Signs: ++ for x<−1x < -1, −- on (−1,1)(-1, 1), ++ for x>1x > 1.
    3. So (−1,2)(-1, 2) is a local max and (1,−2)(1, -2) is a local min.
    4. y′′=6xy'' = 6x: negative for x<0x < 0, positive for x>0x > 0, so (0,0)(0, 0) is a point of inflection.

    Answer: Local max (−1,2)(-1, 2), local min (1,−2)(1, -2), inflection (0,0)(0, 0), xx-intercepts 0,±30, \pm\sqrt{3}.

  2. Medium

    Sketch y=x4−4x3y = x^4 - 4x^3. Find the critical numbers, local extrema and points of inflection.
    Show solution
    1. Intercepts: x3(x−4)=0x^3(x - 4) = 0 gives x=0,4x = 0, 4; the yy-intercept is 00.
    2. y′=4x3−12x2=4x2(x−3)y' = 4x^3 - 12x^2 = 4x^2(x - 3), zero at x=0x = 0 and x=3x = 3.
    3. Signs of y′y': −- for x<0x < 0, −- on (0,3)(0, 3), ++ for x>3x > 3. No sign change at 00, so it is not an extremum. Local min at (3,−27)(3, -27).
    4. y′′=12x2−24x=12x(x−2)y'' = 12x^2 - 24x = 12x(x - 2): ++, then −- on (0,2)(0, 2), then ++. Concavity changes at x=0x = 0 and x=2x = 2.
    5. Points of inflection: (0,0)(0, 0) and (2,−16)(2, -16).

    Answer: Critical numbers 00 and 33; local min (3,−27)(3, -27), no local max; inflection points (0,0)(0, 0) and (2,−16)(2, -16).

  3. Medium

    Sketch y=4xx2+4y = \frac{4x}{x^2 + 4}.
    Show solution
    1. The denominator is never zero, so there is no vertical asymptote. Degree 1<21 < 2, so the horizontal asymptote is y=0y = 0. The function is odd and passes through (0,0)(0, 0).
    2. Quotient rule: y′=4(x2+4)−4x(2x)(x2+4)2=4(4−x2)(x2+4)2y' = \frac{4(x^2 + 4) - 4x(2x)}{(x^2 + 4)^2} = \frac{4(4 - x^2)}{(x^2 + 4)^2}, zero at x=±2x = \pm 2.
    3. Signs of y′y': −-, then ++ on (−2,2)(-2, 2), then −-. Local min (−2,−1)(-2, -1), local max (2,1)(2, 1).
    4. y′′=8x(x2−12)(x2+4)3y'' = \frac{8x(x^2 - 12)}{(x^2 + 4)^3}, zero at x=0,±23x = 0, \pm 2\sqrt{3}, and its sign changes at each.
    5. Points of inflection: (0,0)(0, 0), (23,32)\left(2\sqrt{3}, \frac{\sqrt{3}}{2}\right) and (−23,−32)\left(-2\sqrt{3}, -\frac{\sqrt{3}}{2}\right).

    Answer: HA y=0y = 0; local min (−2,−1)(-2, -1), local max (2,1)(2, 1); inflection points at x=0,±23≈±3.46x = 0, \pm 2\sqrt{3} \approx \pm 3.46.

  4. Test level

    Sketch y=x2x−1y = \frac{x^2}{x - 1}, including all asymptotes.
    Show solution
    1. Vertical asymptote x=1x = 1: the denominator is zero there and the numerator is not.
    2. Numerator degree is one higher, so divide: x2x−1=x+1+1x−1\frac{x^2}{x - 1} = x + 1 + \frac{1}{x - 1}. The oblique asymptote is y=x+1y = x + 1.
    3. y′=2x(x−1)−x2(x−1)2=x(x−2)(x−1)2y' = \frac{2x(x - 1) - x^2}{(x - 1)^2} = \frac{x(x - 2)}{(x - 1)^2}, zero at x=0x = 0 and x=2x = 2.
    4. Signs of y′y': ++ for x<0x < 0, −- on (0,1)(0, 1) and (1,2)(1, 2), ++ for x>2x > 2. Local max (0,0)(0, 0), local min (2,4)(2, 4).
    5. y′′=2(x−1)3y'' = \frac{2}{(x - 1)^3}: concave down for x<1x < 1, concave up for x>1x > 1. It is never zero, so there is no point of inflection.

    Answer: VA x=1x = 1, OA y=x+1y = x + 1; local max (0,0)(0, 0), local min (2,4)(2, 4); concave down left of x=1x = 1, up to the right.

  5. Test level

    Sketch y=x53−5x23y = x^{\frac{5}{3}} - 5x^{\frac{2}{3}}.
    Show solution
    1. Intercepts: x23(x−5)=0x^{\frac{2}{3}}(x - 5) = 0 gives x=0x = 0 and x=5x = 5.
    2. y′=53x23−103x−13=5(x−2)3x13y' = \frac{5}{3}x^{\frac{2}{3}} - \frac{10}{3}x^{-\frac{1}{3}} = \frac{5(x - 2)}{3x^{\frac{1}{3}}}. Zero at x=2x = 2, undefined at x=0x = 0.
    3. Signs of y′y': ++ for x<0x < 0, −- on (0,2)(0, 2), ++ for x>2x > 2. Local max at (0,0)(0, 0), a cusp, since y′y' is undefined there. Local min at (2,−343)≈(2,−4.76)(2, -3\sqrt[3]{4}) \approx (2, -4.76).
    4. y′′=109x−13+109x−43=10(x+1)9x43y'' = \frac{10}{9}x^{-\frac{1}{3}} + \frac{10}{9}x^{-\frac{4}{3}} = \frac{10(x + 1)}{9x^{\frac{4}{3}}}. The denominator is positive for x≠0x \ne 0, so the sign follows x+1x + 1.
    5. Concave down for x<−1x < -1, concave up on (−1,0)(-1, 0) and x>0x > 0. Point of inflection at (−1,−6)(-1, -6).

    Answer: Cusp and local max at (0,0)(0, 0), local min (2,−343)(2, -3\sqrt[3]{4}), inflection point (−1,−6)(-1, -6), xx-intercepts 00 and 55.

Learn It on Martian Lab

Martian Lab teaches this in the Curve Sketching lesson of its MCV4U Curve Sketching unit. Watch it, then take the Curve Sketching practice tests to see if it stuck.

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