Linear growth and decay

In \(Q=80-5t\), the coefficient of \(t\) is the constant \(-5\), so the quantity falls by 5 units per time interval. \(80+5t\) shows growth, not decay. Exam tip: linear decay has a constant negative slope.

In \(y=-3x+12\), the coefficient of \(x\) is \(-3\), so \(y\) falls by 3 units for every increase of 1 in \(x\). Substituting \(x=0\) gives 12, which is positive. Option C has a negative initial value. Exam tip: decay requires a negative slope.

For the models to be equal, set \(45+8t=129-4t\). Moving \(4t\) to the left and \(45\) to the right gives \(12t=84\). Hence, \(t=7\), so option C is correct. \(t=6\) is the nearest distractor, but the two model values are not equal at that time. Exam tip: Equate the two expressions and collect all variable terms on one side.

For equality, set the two expressions equal: \(38+6t=122-8t\). Bringing \(8t\) to the left gives \(14t=84\), so \(t=6\). Although \(t=8\) is a close distractor, \(P(8)=86\) and \(Q(8)=58\), so they are not equal. Exam tip: when one expression increases and the other decreases, equate them to find their intersection time.

Given \(f(2)=a+11\times2=61\), we get \(a+22=61\), so \(a=39\). Therefore, \(f(9)=39+11\times9=39+99=138\). Option 127 may result from incorrectly calculating or adding the increase \(11\times(9-2)=77\). Exam tip: In a linear function, multiply the change in \(x\) by its coefficient to find the change in the function value quickly.

Given \(g(x)=b-7x\), we have \(g(6)=b-7(6)=b-42\). Since \(g(6)=74\), \(b-42=74\), so \(b=116\). Now, \(g(13)=116-7(13)=116-91=25\). Therefore, 25 is the correct option. A value such as 32 can result from an error in subtraction. Exam tip: first find the unknown constant from the given function value, then substitute the required value of \(x\).

In \(y=12-3x\), the constant term 12 gives a positive initial value, while the coefficient −3 makes y decrease as x increases. \(y=12+3x\) represents growth instead. Exam tip: check the signs of the intercept and slope separately.

In \(y=18-3x\), the coefficient of \(x\) is a constant negative number, so \(y\) decreases by 3 units for every 1-unit increase in \(x\). Option B shows growth, while C and D are non-linear. Exam tip: in \(y=a+bx\), decay occurs when \(b<0\).

In \(y=40-3x\), the highest power of \(x\) is 1, and the coefficient \(-3\) means \(y\) decreases by 3 when \(x\) increases by 1. \(40-3x^2\) is quadratic. Exam tip: a linear polynomial has degree 1.

\(12-3x\) is a first-degree polynomial with coefficient \(-3\) of \(x\). Thus, the quantity decreases by 3 for every increase of 1 in \(x\). \(12-3x^2\) is not linear because its rate changes. Exam tip: linear decay has a negative coefficient of \(x\).

Given \(C(t)=31+kt\), we have \(C(9)=31+9k\) and \(C(4)=31+4k\). Hence, \(C(9)-C(4)=(31+9k)-(31+4k)=5k\). Since this difference is 60, \(5k=60\), so \(k=12\). Option 15 would result from incorrectly taking the time difference as 4; the correct difference is \(9-4=5\). Exam tip: when subtracting two values of a linear expression, the constant term cancels out.

Given \(D(t)=190-kt\), we have \(D(3)=190-3k\) and \(D(12)=190-12k\). Therefore, \(D(3)-D(12)=(190-3k)-(190-12k)=9k\). Since \(9k=81\), \(k=9\). Option 12 is the difference between the time values, not the value of \(k\). Exam tip: Write both function values separately before subtracting; the constant terms often cancel out.

Here, \(A(x+6)=25+4(x+6)=49+4x\) and \(A(x-1)=25+4(x-1)=21+4x\). Therefore, \(A(x+6)-A(x-1)=(49+4x)-(21+4x)=28\). Hence, 28 is correct. Option 24 would result from using a change of only 6, but the actual difference between \((x+6)\) and \((x-1)\) is 7. Exam tip: for a linear function \(a+bx\), multiply the change in input by \(b\) to quickly find the change in output.

In \(P(x)=180-4x\), the constant term 180 gives the initial value, while the coefficient \(-4\) shows a decrease of 4 per unit increase in \(x\). Option A represents growth. Exam tip: a negative coefficient of \(x\) indicates linear decay.

Given P(t)=p+9t, we have P(3)=p+9(3)=82. Thus p+27=82, so p=55. Now P(s)=55+9s=136, which gives 9s=81 and hence s=9. If 8 were used, the value would be 127, not 136. Exam tip: first find the constant term p from the known input-output pair, then substitute it into the second condition.

Given \(R(t)=r-8t\), substitute \(R(5)=104\): \(r-8(5)=104\), so \(r=144\). Now \(R(s)=48\) gives \(144-8s=48\). Thus \(8s=96\), and \(s=12\). For example, \(s=14\) would not give a value of 48. Exam tip: first find the constant \(r\) from the known input-output pair, then use the second condition.

An increase of 8 per step requires a linear expression with slope 8, so \(y=a+8x\). Using \(y(3)=50\) gives \(a+8(3)=50\), hence \(a=26\). Therefore, \(y=26+8x\) is correct. Although \(y=74-8x\) gives \(y(3)=50\), its slope is \(-8\), so it represents decay rather than growth. Exam tip: check both the slope and the given point.

A decrease of 4 per step means the coefficient of \(x\) must be \(-4\). So the equation has the form \(y=a-4x\). Using \(y(7)=73\), we get \(a-4(7)=73\), so \(a=101\). Therefore, \(y=101-4x\) is correct. In option A, the initial value is 73, giving \(y(7)=45\), not 73. Exam tip: substitute the given \(x\)-value into each suitable option to check the condition quickly.

“14 more” means \(M(t)=N(t)+14\). Thus, \(58+3t=116-5t+14=130-5t\). Hence \(8t=72\), so \(t=9\). At \(t=8\), the difference is only \(6\), making it a close but incorrect option. Exam tip: for “more than” questions, write the larger quantity as the smaller quantity plus the stated difference.

“2 more than \(B(t)\)” means \(A(t)=B(t)+2\). So, \(170-9t=38+4t+2\), which gives \(170-9t=40+4t\) and \(130=13t\). Hence, \(t=10\). At \(t=9\), the difference is 15, not 2. Exam tip: In comparison questions, first convert the stated difference into an equation.