The graphs below represent the functions defined by
g(x) = - (x + 2)(x - 1)(x - 3) and
h(x) = 2x + p
E and F are the turning points of g - NSC Technical Mathematics - Question 7 - 2022 - Paper 1
Question 7
The graphs below represent the functions defined by
g(x) = - (x + 2)(x - 1)(x - 3) and
h(x) = 2x + p
E and F are the turning points of g.
A, B, C and D are i... show full transcript
Worked Solution & Example Answer:The graphs below represent the functions defined by
g(x) = - (x + 2)(x - 1)(x - 3) and
h(x) = 2x + p
E and F are the turning points of g - NSC Technical Mathematics - Question 7 - 2022 - Paper 1
Step 1
Write down the coordinates of C.
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Answer
The coordinates of point C can be derived from the graph where the curve intersects the x-axis at point C. Thus, the coordinates of C are (3; 0).
Step 2
Write down the value of p.
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Answer
To find the value of p, we use the function h(x) = 2x + p. At point C where x = 3, h(3) = 0.
Setting up the equation:
0 = 2(3) + p
0 = 6 + p
Solving for p gives us:
p = -6.
Step 3
Determine the length of AC.
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Answer
To find the length of segment AC on the graph, we use the coordinates of A (0, 0) and C (3, 0).
Using the distance formula:
Length AC = |x_C - x_A| = |3 - 0| = 3 units. Thus the final answer:
Length AC = 3 units.
Step 4
Express g(x) = - (x + 2)(x - 1)(x - 3) in the form g(x) = ax³ + bx² + cx + d.
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Answer
To express g(x) in standard polynomial form:
Start with expansion:
g(x) = -((x + 2)(x - 1)(x - 3))
First multiply (x - 1)(x - 3):
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Answer
E and F are turning points obtained by finding the derivative of g(x) and setting it to zero.
The derivative of g(x) is:
g'(x) = -3x² + 2(2x) + 5.
Setting g'(x) = 0, solve for x values which gives E(0, -8.21) and F(2.12, 4.06).
Step 6
Write down the values of x for which g(x) > 0.
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Answer
To determine the values of x for which g(x) > 0, we must find the x-values between the roots of the function. Inspecting the graph, it can be concluded that g(x) > 0 when x ∈ (-∞, -2) and (1, 3).
