Die grafiek van $h(x)=ax^3+bx^2$ is gesketst - NSC Mathematics - Question 10 - 2021 - Paper 1
Question 10
Die grafiek van $h(x)=ax^3+bx^2$ is gesketst.
Die grafiek het draainpunt by die oorsprong, $O(0; 0)$ en $B(4; 32)$.
A is 'n $x$-afsnit van $h$.
10.1 Toon dat ... show full transcript
Worked Solution & Example Answer:Die grafiek van $h(x)=ax^3+bx^2$ is gesketst - NSC Mathematics - Question 10 - 2021 - Paper 1
Step 1
Toon dat $a=-1$ en $b=6$
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Answer
We begin by computing the first derivative of the function to find critical points.
Given: h′(x)=3ax2+2bx
Substituting x=4 where h(4)=32: h(4)=64a+16b=32
Next, substituting into our equation gives us: 4a+b=2ext(1)
Using the first derivative for h′(4) and solving:
3a(4^2) + 2b(4) = 0 \
48a + 8b = 0\ ext{ (2)}$$
Solving equations (1) and (2) simultaneously:
From (1), rearranging gives $b = 2 - 4a$. Substituting this into (2):
$$48a + 8(2 - 4a) = 0 \
48a + 16 - 32a = 0 \
16a = -16 \
a = -1 \ $$
By substituting back we find:
$$b = 2 - 4(-1) = 6.$$
Step 2
Bereken die koördinate van A.
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Answer
To find the coordinates of point A where x=0, we substitute x=0 into the function: h(0)=a(0)3+b(0)2=0.
Therefore, the coordinates of A are (0,0).
Step 3
Skryf die waarde van $x$ neer waarvoor: h' is 0.
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Answer
The critical values occur when h′=0: 0=3(−1)x2+2(6)x
This simplifies to:
x(12 - 3x) = 0$$
Thus, the values of $x$ are $x = 0$ and $x = 4$.
Step 4
Konkafiteit is +.
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Answer
The concavity is determined by the second derivative: h′′(x)=6ax+2b.
Substituting in our values for a and b: h′′(x)=6(−1)x+12.
Set h′′(x)>0 for concavity to be positive:
-6x + 12 > 0 \
x < 2.$$
Therefore, $0 ext{ up to } 2$ is where concavity is positive.
Step 5
Vir watter waarde$s$ van $k$ sal $-(x-1)^3 + 6(x-1)^2 - k = 0$ een negatief en twee verskillende positiewe wortels hê?
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Answer
To analyze the cubic equation, we can utilize Descartes' Rule of Signs for root counting.
We are interested in −(x−1)3+6(x−1)2−k=0.
For a cubic function to have one negative and two positive roots, the discriminant must be positive. The condition for two distinct positive roots is met when k<32.
Thus, the values of k must be such that: k<32.
