6.1 Write down the equation of $g^{-1}$ in the form $y = ...$ - NSC Mathematics - Question 6 - 2021 - Paper 1
Question 6
6.1 Write down the equation of $g^{-1}$ in the form $y = ...$.
6.1.2 Point $P(6 ; 11)$ lies on $h(x) = 3^{x-4} + 2$. The graph of $h$ is translated to form $g$. Wri... show full transcript
Worked Solution & Example Answer:6.1 Write down the equation of $g^{-1}$ in the form $y = ...$ - NSC Mathematics - Question 6 - 2021 - Paper 1
Step 1
Write down the equation of $g^{-1}$ in the form $y = ...$
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Answer
To find the equation of g−1, we need to swap x and y in the equation of g(x)=3x.
Hence, the equation becomes:
y=extlog3x.
Step 2
Point $P(6 ; 11)$ lies on $h(x) = 3^{x-4} + 2$. The graph of $h$ is translated to form $g$. Write down the coordinates of the image of $P$ on $g$.
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Answer
Point P(6;11) has coordinates:
The x-coordinate of P is translated 4 units left, yielding x=6−4=2.
The y-coordinate is translated 2 units down, yielding y=11−2=9.
Thus, the coordinates of the image of P on g are (2;9).
Step 3
Determine the values of $p$ and $q$
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Answer
Given the function f(x)=2p+q, we know that the asymptote is at y=−16, which implies:
q=−16.
Substituting the point T(3;16) into the function, we get:
16=2p+q.
Substituting q=−16:
16=2p−16.
Thus:
2p=32.
We can express 32 as a power of 2:
ightarrow p = 5.$$
Hence, the values are:
- $p = 5$
- $q = -16$.
