Four linked problems drawn from this topic’s subtopics. 12 parts · untimed · answers and worked solutions at the end.
Example 1
The curve is f(x)=aln(x−h)+k, with vertical asymptote x=−1 and domain x>−1. The graph shows a finite window of this full curve. The labelled point coordinates are exact; the horizontal coordinate −1+e is not a rounded measurement. Use the graph to recover the constants.
Use the two labelled points to determine a.
y
0123-7.5-5-2.50AB
x
A: (0,−3)B: (−1+e,−1)
fShow worked answer
Answer: 2
At x=0,ln(x−h)=0, so the plotted value gives k=−3.
At x=−1+e, the logarithm equals 1, so a+k=−1. Therefore a=2.
Example 2
The graph plots log2P against x, where P>0. It is a straight line for all real x. The two labelled coordinates are exact.
a Write the line as log2P=mx+b. Find m.
Find the exact initial value P(0).
log2(P)
012340246AB
x
A: (1,2)B: (3,4)
log2(P)Show worked answer
Answer: 2
log2P(0)=1. Invert the logarithm: P(0)=21.
Thus P(0)=2.
Example 3
The graph plots log2P against x, where P>0. It is a straight line for all real x. The two labelled coordinates are exact.
a Write the line as log2P=mx+b. Find m.
b Find the exact initial value P(0).
A second model is Q(x)=24x+1. Find the exact x−value where P(x)=Q(x).
log2(P)
01234-50510AB
x
A: (1,2)B: (3,8)
log2(P)Show worked answer
Answer: −2
The graph represents P(x)=23x−1. Equal powers of the same base greater than 1 have equal exponents.
Therefore 3x−1=4x+1, so x=−2.
For inputs above this crossing, P(x)<Q(x) because the difference between their exponents has slope −1.
Differential Calculus Review Practice | Year 12 Methods | Ascent Arcade