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7 (a) Sketch the graph of any cubic function that has both three distinct real roots and a positive coefficient of $x^3$ - AQA - A-Level Maths Mechanics - Question 7 - 2019 - Paper 2

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7 (a) Sketch the graph of any cubic function that has both three distinct real roots and a positive coefficient of $x^3$. 7 (b) The function $f(x)$ is defined by ... show full transcript

Worked Solution & Example Answer:7 (a) Sketch the graph of any cubic function that has both three distinct real roots and a positive coefficient of $x^3$ - AQA - A-Level Maths Mechanics - Question 7 - 2019 - Paper 2

Step 1

Sketch the graph of any cubic function that has both three distinct real roots and a positive coefficient of $x^3$

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Answer

To sketch the graph of a cubic function with three distinct real roots and a positive leading coefficient, we note that the graph will have the following characteristics:

  1. Positive Leading Coefficient: The end behavior of the cubic function will rise to the right and rise to the left.

  2. Distinct Real Roots: The graph will cross the x-axis at three points, indicating three distinct real roots, denoted as r1r_1, r2r_2, and r3r_3. The graph will change direction at each root.

  3. Sketch Representation:

    • Start with three x-intercepts (r1r_1, r2r_2, r3r_3) marked on the x-axis.
    • The graph typically will dive down from the left, cross the x-axis at r1r_1, rise upward, cross again at r2r_2, and finally fall, crossing at r3r_3 before rising off to the right.

This results in a graph that smoothly curves up and down, representing the positive coefficient and the three distinct real roots.

Step 2

Show that there is a turning point where the curve crosses the $y$-axis.

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Answer

To show that there is a turning point where the curve crosses the yy-axis, we start with the given function:

f(x)=x3+3px2+qf(x) = x^3 + 3px^2 + q

  1. Differentiate the Function: f(x)=3x2+6pxf'(x) = 3x^2 + 6px
    We need to find the turning points where f(x)=0f'(x) = 0.

  2. Set Derivative to Zero: 3x2+6px=03x^2 + 6px = 0
    Factor out common terms: 3x(x+2p)=03x(x + 2p) = 0 This gives us the solutions:

    • x=0x = 0
    • x=2px = -2p
  3. Analyzing the Turning Points:

    • Since p>0p > 0, it follows that 2p<0-2p < 0.
    • We have established turning points at x=0x = 0 and x=2px = -2p.
  4. Evaluate the Function at x=0x = 0: f(0)=qf(0) = q
    This indicates that when the curve crosses the yy-axis (at x=0x = 0), the value of the function is f(0)=qf(0) = q.

  5. Determine the Nature of the Turning Point at x=0x = 0:

    • At x=0x = 0, if f(0)=0f'(0) = 0 and f(0)=qf(0) = q, then this point can act as a minimum or maximum.

Since we can see that there is a turning point at x=0x=0 where the curve crosses the yy-axis, we have shown that such a point exists.

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