Porisma

Review practice · Section 1 of 2

Questions 1–6 in this collection

Sketch a derivative across two vertical asymptotes · Sketch the gradient across a pole and a minimum · +4 more

All chapter practiceAll review practice
  1. 1

    The diagram, not to scale, shows the graph of y=f(x)y=f(x). The vertical asymptotes are x=0x=0 and x=2x=2. Sketch the graph of y=f′(x)y=f^{\prime}(x), showing all corresponding coordinates and asymptotes where possible. marks

    Expand diagram

    Use the scrollbars or arrow keys to inspect details.

    Qualitative graph, not to scale: local maximum A(-1,2) on the negative-x branch, zero at -0.5, vertical asymptote x=0; between 0 and 2 a U-shaped branch has minimum B(1,3); beyond x=2 the curve rises from negative infinity towards the x-axis. No numerical function is supplied.
    Qualitative graph, not to scale: local maximum A(-1,2) on the negative-x branch, zero at -0.5, vertical asymptote x=0; between 0 and 2 a U-shaped branch has minimum B(1,3); beyond x=2 the curve rises from negative infinity towards the x-axis. No numerical function is supplied.
    Qualitative graph, not to scale: local maximum A(-1,2) on the negative-x branch, zero at -0.5, vertical asymptote x=0; between 0 and 2 a U-shaped branch has minimum B(1,3); beyond x=2 the curve rises from negative infinity towards the x-axis. No numerical function is supplied.

    [2010/SRJC/I/6(a)]

    CJC · Review problems · Question 1

  2. 2

    The graph of y=f(x)y=f(x) has asymptotes x=2x=2, y=−3y=-3 and minimum (−2,−5)(-2,-5), as shown. Sketch y=f′(x)y=f^{\prime}(x), clearly indicating any asymptotes, turning points and axial intercepts. marks

    The supplied graph also labels x-intercepts 1,41,4 and y-intercept −3-3.

    Expand diagram

    Use the scrollbars or arrow keys to inspect details.

    Given function with vertical asymptote x=2 and horizontal asymptote y=−3. The left branch initially falls below −3, reaches minimum (−2,−5), then rises through (0,−3) and (1,0) to positive infinity at 2 from the left. The right branch falls from positive infinity at 2 from the right, crosses (4,0), then tends to −3 from above.
    Given function with vertical asymptote x=2 and horizontal asymptote y=−3. The left branch initially falls below −3, reaches minimum (−2,−5), then rises through (0,−3) and (1,0) to positive infinity at 2 from the left. The right branch falls from positive infinity at 2 from the right, crosses (4,0), then tends to −3 from above.
    Given function with vertical asymptote x=2 and horizontal asymptote y=−3. The left branch initially falls below −3, reaches minimum (−2,−5), then rises through (0,−3) and (1,0) to positive infinity at 2 from the left. The right branch falls from positive infinity at 2 from the right, crosses (4,0), then tends to −3 from above.

    [2009/CJC/Promo/10(c)]

    CJC · Review problems · Question 2

  3. 3

    The diagram shows y=f(x)y=f(x) with a maximum at x=ax=a, minimum at x=bx=b, and stationary point of inflection at the origin. On a separate diagram draw y=f′(x)y=f^{\prime}(x). marks

    Expand diagram

    Use the scrollbars or arrow keys to inspect details.

    Given function rises to a maximum at negative x=a, decreases through a stationary inflection at the origin, continues to a minimum at positive x=b, then rises. The two turning-point x-values are marked a and b without numerical coordinates.
    Given function rises to a maximum at negative x=a, decreases through a stationary inflection at the origin, continues to a minimum at positive x=b, then rises. The two turning-point x-values are marked a and b without numerical coordinates.
    Given function rises to a maximum at negative x=a, decreases through a stationary inflection at the origin, continues to a minimum at positive x=b, then rises. The two turning-point x-values are marked a and b without numerical coordinates.

    [2009/SRJC/Promo/2]

    CJC · Review problems · Question 3

  4. 4

    The diagram shows y=f(x)y=f(x) with asymptotes x=3x=3 and y=−x+2y=-x+2, y-intercept 7/37/3, minimum (2,1)(2,1) and maximum (4,−3)(4,-3).

    Sketch the curve y=f′(x)y=f^{\prime}(x). marks

    Expand diagram

    Use the scrollbars or arrow keys to inspect details.

    Given function with asymptotes x=3 and y=−x+2, y-intercept 7/3, minimum (2,1) on the left branch and maximum (4,−3) on the right. The left branch tends up without bound at 3 from the left; the right starts at negative infinity and rises to its maximum before falling.
    Given function with asymptotes x=3 and y=−x+2, y-intercept 7/3, minimum (2,1) on the left branch and maximum (4,−3) on the right. The left branch tends up without bound at 3 from the left; the right starts at negative infinity and rises to its maximum before falling.
    Given function with asymptotes x=3 and y=−x+2, y-intercept 7/3, minimum (2,1) on the left branch and maximum (4,−3) on the right. The left branch tends up without bound at 3 from the left; the right starts at negative infinity and rises to its maximum before falling.

    [2010/VJC/Promo/6(ii)]

    CJC · Review problems · Question 4

  5. 5

    The diagram shows y=f′(x)y=f^{\prime}(x) meeting the x-axis at x=−3x=-3 and x=1x=1. Given f(−3)=4f(-3)=4 and f(1)=1f(1)=1:

    (i) State the coordinates and natures of the stationary points of y=f(x)y=f(x). marks

    (ii) Find the range of xx for which the curve is strictly decreasing. marks

    (iii) Sketch y=f(x)y=f(x). marks

    Expand diagram

    Use the scrollbars or arrow keys to inspect details.

    Given derivative is negative before −3, touches zero at −3 as a local maximum, falls to an unlabelled negative minimum, then rises and crosses the axis at 1. It is positive after 1.
    Given derivative is negative before −3, touches zero at −3 as a local maximum, falls to an unlabelled negative minimum, then rises and crosses the axis at 1. It is positive after 1.
    Given derivative is negative before −3, touches zero at −3 as a local maximum, falls to an unlabelled negative minimum, then rises and crosses the axis at 1. It is positive after 1.

    [2010/TPJC/I/6]

    CJC · Review problems · Question 5

  6. 6

    The curve CC is y=f(x)=ax2+bx+cy=f(x)=\dfrac{ax^2+b}{x+c}, where a,b,ca,b,c are constants, and y=−x+2y=-x+2 is an asymptote.

    (i) Show that a=−1a=-1, c=2c=2. marks

    (ii) Given b=1b=1, sketch CC, showing all asymptotes, stationary points and axial intercepts, if any. marks

    (iii) Sketch y=f′(x)y=f^{\prime}(x), showing all asymptotes, axial intercepts and stationary points, if any. marks

    [2010/TPJC/Promo/10(i),(ii),(iii)(a)]

    CJC · Review problems · Question 6

Back to all questions
Choose another collectionChapter practice