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
Question 7
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:
Positive Leading Coefficient: The end behavior of the cubic function will rise to the right and rise to the left.
Distinct Real Roots: The graph will cross the x-axis at three points, indicating three distinct real roots, denoted as r1, r2, and r3. The graph will change direction at each root.
Sketch Representation:
Start with three x-intercepts (r1, r2, r3) marked on the x-axis.
The graph typically will dive down from the left, cross the x-axis at r1, rise upward, cross again at r2, and finally fall, crossing at r3 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 y-axis, we start with the given function:
f(x)=x3+3px2+q
Differentiate the Function:
f′(x)=3x2+6px
We need to find the turning points where f′(x)=0.
Set Derivative to Zero:
3x2+6px=0
Factor out common terms:
3x(x+2p)=0
This gives us the solutions:
x=0
x=−2p
Analyzing the Turning Points:
Since p>0, it follows that −2p<0.
We have established turning points at x=0 and x=−2p.
Evaluate the Function at x=0:
f(0)=q
This indicates that when the curve crosses the y-axis (at x=0), the value of the function is f(0)=q.
Determine the Nature of the Turning Point at x=0:
At x=0, if f′(0)=0 and f(0)=q, then this point can act as a minimum or maximum.
Since we can see that there is a turning point at x=0 where the curve crosses the y-axis, we have shown that such a point exists.