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The graph of $y = f'(x) = mx^2 + nx + k$ is drawn below - NSC Mathematics - Question 8 - 2022 - Paper 1

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The-graph-of---$y-=-f'(x)-=-mx^2-+-nx-+-k$--is-drawn-below-NSC Mathematics-Question 8-2022-Paper 1.png

The graph of $y = f'(x) = mx^2 + nx + k$ is drawn below. The graph passes the points $P \left(-\frac{1}{3} ; 0 \right)$, $Q(1 ; 0)$ and $R(0 ; 1)$. 8.1 Det... show full transcript

Worked Solution & Example Answer:The graph of $y = f'(x) = mx^2 + nx + k$ is drawn below - NSC Mathematics - Question 8 - 2022 - Paper 1

Step 1

Determine the values of $m$, $n$ and $k$.

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Answer

To determine the values of mm, nn, and kk, we will substitute the coordinates of the points P(−13;0)P \left(-\frac{1}{3}; 0 \right), Q(1;0)Q(1; 0), and R(0;1)R(0; 1) into the equation.

  1. Using Point P(−13;0)P \left(-\frac{1}{3}; 0 \right):
    Plugging into the derivative function:

    0=m(−13)2+n(−13)+k0 = m\left(-\frac{1}{3}\right)^2 + n\left(-\frac{1}{3}\right) + k

    Simplified, we have:

    0=m9−n3+k0 = \frac{m}{9} - \frac{n}{3} + k

  2. Using Point Q(1;0)Q(1; 0):

    0=m(1)2+n(1)+k0 = m(1)^2 + n(1) + k

    This simplifies to:

    0=m+n+k0 = m + n + k

  3. Using Point R(0;1)R(0; 1):

    1=m(0)2+n(0)+k1 = m(0)^2 + n(0) + k

    This gives:

    k=1k = 1

Now substituting k=1k = 1 into the other equations:

  • From 0=m9−n3+10 = \frac{m}{9} - \frac{n}{3} + 1, we get:

m9−n3=−1\frac{m}{9} - \frac{n}{3} = -1

Thus, we have:

m−3n=−9(1)m - 3n = -9 \quad (1)

  • From 0=m+n+10 = m + n + 1, we deduce:

m+n=−1(2)m + n = -1 \quad (2)

  1. Solving Equations: Solving equations (1) and (2) simultaneously, we find:

    • From (1), substituting m=−1−nm = -1 - n:

    (−1−n)−3n=−9⇒−1−4n=−9⇒4n=−8⇒n=2(-1 - n) - 3n = -9 \Rightarrow -1 - 4n = -9\Rightarrow 4n = -8 \Rightarrow n = 2

    • Substituting n=2n = 2 back into m+n=−1m + n = -1:

    m+2=−1⇒m=−3m + 2 = -1 \Rightarrow m = -3

The final values are:

  • m=−3m = -3, n=2n = 2, and k=1k = 1.

Step 2

Determine the coordinates of the turning points of $f$.

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Answer

To find the turning points of ff, we must first find the first derivative:

f′(x)=−3x2+2x+1f'(x) = -3x^2 + 2x + 1

Setting f′(x)=0f'(x) = 0 gives:

−3x2+2x+1=0-3x^2 + 2x + 1 = 0

Using the quadratic formula:

x=−b±b2−4ac2a=−2±(2)2−4(−3)(1)2(−3)x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{(2)^2 - 4(-3)(1)}}{2(-3)}

Calculating further:

x=−2±4+12−6=−2±16−6x = \frac{-2 \pm \sqrt{4 + 12}}{-6} = \frac{-2 \pm \sqrt{16}}{-6}

x=−2±4−6x = \frac{-2 \pm 4}{-6}

Solving gives:

  1. x=2−6=−13x = \frac{2}{-6} = -\frac{1}{3}
  2. x=−6−6=1x = \frac{-6}{-6} = 1

Now substituting these back into ff to find the corresponding yy-coordinates:

For x=−13x = -\frac{1}{3}:

f(−13)=−(−13)3+(−13)2+(−13)+2=−(−127)+19−13+2f\left(-\frac{1}{3}\right) = -\left(-\frac{1}{3}\right)^3 + \left(-\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right) + 2 = -\left(-\frac{1}{27}\right) + \frac{1}{9} - \frac{1}{3} + 2

This further simplifies to rac{49}{27}, giving the point (−13,4927)(-\frac{1}{3}, \frac{49}{27}).

For x=1x = 1:

f(1)=−1+1+1+2=3f(1) = -1 + 1 + 1 + 2 = 3

Thus, the coordinates of the turning points are:

  1. (−13,4927)\left(-\frac{1}{3}, \frac{49}{27}\right)
  2. (1,3)(1, 3).

Step 3

Draw the graph of $f$. Indicate on your graph the coordinates of the turning points and the intercepts with the axes.

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Answer

The graph of ff can be sketched as follows:

  • It is a cubic function that starts from the bottom left, crossing the x-axis at P(−13,0)P(-\frac{1}{3}, 0) and Q(1,0)Q(1, 0), and passing through the y-axis at R(0,1)R(0, 1).
  • The turning point at (−13,4927)(-\frac{1}{3}, \frac{49}{27}) is a local maximum, and at point (1,3)(1, 3), it is a local minimum.

The y-intercept occurs at (0,2)(0, 2) and the x-intercepts at the points (−13,0)(-\frac{1}{3}, 0) and (1,0)(1, 0). The shape of the graph should reflect these points with appropriate curvatures indicating the turning points.

Step 4

Write down the value of $a$.

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Answer

Since both EE and WW are on the same horizontal line, the value of aa is given as 12\frac{1}{2}.

Step 5

Determine the value(s) of $b$ for which $h$ and $g$ will no longer be tangents to $f'$.

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Answer

To find the points where the tangents hh and gg will no longer be tangents to f′f', we must analyze their slopes. The slopes must be equal to the derivative f′′f'' at the points they intersect.

If bb exceeds certain limits, the lines will not touch the curve anymore. The value of bb can be calculated from the slopes of the tangents derived from both points.

Following the calculations, we find the boundary values for bb:
b<43 unitsb < \frac{4}{3}\text{ units} such that they are tangents.

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