Geteken hieronder is 'n skets van die hiperbol f(x) = \frac{d - x}{x - p}, waar p en d konstantes is.\n\nDie stippeleine is die asymptote - NSC Mathematics - Question 7 - 2017 - Paper 1
Question 7
Geteken hieronder is 'n skets van die hiperbol f(x) = \frac{d - x}{x - p}, waar p en d konstantes is.\n\nDie stippeleine is die asymptote. Die punt A(5; 0) is gegee ... show full transcript
Worked Solution & Example Answer:Geteken hieronder is 'n skets van die hiperbol f(x) = \frac{d - x}{x - p}, waar p en d konstantes is.\n\nDie stippeleine is die asymptote - NSC Mathematics - Question 7 - 2017 - Paper 1
Step 1
7.1 Bepaal die waardes van d en p.
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Answer
To determine the values of d and p, we observe the point A(5; 0) on the graph. Given that A is on the hyperbola, we can substitute x=5 and f(x)=0 into the equation:
0=5−pd−5
This leads to:
d−5=0
Thus, we find that d=5.
Next, substituting d=5 into the equation:
0=5−p5−5
The above will hold true if p=2. Therefore, the values are:
d=5
p=2
Step 2
7.2 Toon aan dat die vergelyking geskryf kan word as \frac{3}{x - 2} - 1.
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Answer
To rewrite the equation, we start with the original form:
y=x−25−x
To manipulate this expression, we can break it down into simpler fractions by performing polynomial long division or simplifying directly:
We can rewrite 5−x as −(x−5), giving:
y=−x−2x−5
Then, by rearranging:
y=x−2−(x−2+3)=−1−x−23
Thus, we have:
y=x−23−1.
Step 3
7.3 Skryf die spieëlbeeld van A, as 'n gerekfekteer word in die simmetrie-as y = x - 3, neer.
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Answer
To find the image of the point A(5; 0) reflected across the line y=x−3, we first find the intersection of the line through A perpendicular to y=x−3.
The slope of y=x−3 is 1, so the slope of the perpendicular line is −1.
The equation of the line through A can be written as:
ightarrow y = -x + 5.$$
3. Setting $y = -x + 5$ equal to $y = x - 3$, we solve for their intersection:
$$-x + 5 = x - 3\
2x = 8 \
x = 4.$$
4. Substituting $x = 4$ back into either equation:
$$y = 4 - 3 = 1.$$
Thus, intersection point is (4; 1).
5. The midpoint of A(5; 0) and its image A'(x, y) can be calculated. If we denote A' as (x, y), then:
$$\left( \frac{5 + x}{2}, \frac{0 + y}{2} \right) = (4, 1).$$
So,
$$\frac{5 + x}{2} = 4 \rightarrow 5 + x = 8 \rightarrow x = 3,$$\n$$\frac{0 + y}{2} = 1 \rightarrow y = 2.$$
Thus the image point A' will be (3; 2).
