Linear growth and decay

Given \(f(x)=8x+22\), we get \(f(3x)=8(3x)+22=24x+22\). Hence, \(f(3x)-f(x)=(24x+22)-(8x+22)=16x\). Setting \(16x=80\) gives \(x=5\). Therefore, \(x=5\) is correct. For \(x=4\), the difference would be only \(64\), not 80. Exam tip: while finding \(f(3x)\), replace every \(x\) in the function with \(3x\).

Given \(g(x)=180-9x\), we get \(g(3x)=180-9(3x)=180-27x\). Hence, \(g(x)-g(3x)=(180-9x)-(180-27x)=18x\). Setting \(18x=90\) gives \(x=5\). Therefore, option B is correct. For \(x=4\), the difference would be only \(72\), not \(90\). Exam tip: while finding \(g(3x)\), substitute the complete expression \(3x\) in place of \(x\).

Given \(P(t)=50+4t\), we get \(P(4)=66\) and \(P(12)=98\). Thus, \(P(4)+P(12)=164\), and half of this is \(82\). Since \(P(8)=50+4\times8=82\), we have \(P(4)+P(12)=2P(8)\). Although \(P(7)=78\) is close, twice it is \(156\), so it is not correct. Exam tip: for a linear function, the function value at the midpoint of two inputs equals the average of the two function values.

This is a linear function. The midpoint of 3 and 15 is \(\frac{3+15}{2}=9\), so for a linear function, \(Q(3)+Q(15)=2Q(9)\). Checking directly, \(Q(3)=189\) and \(Q(15)=105\), giving a sum of \(294\), while \(2Q(9)=2\times147=294\). Hence, Q(9) is correct. Q(10) is not the midpoint, so twice its value is not equal to the given sum. Exam tip: For a linear function, the sum of outputs at two equally spaced inputs equals twice the output at their midpoint.

In \(y=120+ax\), \(a\) is the rate of change of \(y\) with respect to \(x\). The change in \(x\) is \(14-6=8\), so the change in \(y\) is \(8a\). Given \(8a=88\), we get \(a=11\). If \(a=10\), the increase would be \(8\times10=80\), not 88. Exam tip: while finding a change in a linear expression, the constant term \(120\) cancels out.

The change in x is 13 - 5 = 8. In y = 260 - bx, y decreases by b for every increase of 1 in x. Hence, the total decrease over 8 units is 8b. So, 8b = 104, giving b = 104/8 = 13. Option 12 is close, but it would give a total decrease of 8 × 12 = 96, not 104. Exam tip: For a linear relation, use total change = rate per unit × change in x.

The condition is \(A(t)=B(t)+6\). Thus, \(30+5t=54+3t+6\). Simplifying gives \(2t=30\), so \(t=15\). At \(t=14\), \(A(t)-B(t)=4\), not 6. Exam tip: Translate “6 more than” as \(A=B+6\).

The condition is (200-8t=70+4t-10). It gives (140=12t), so (t=\frac{35}{3}), hence no option is correct.

Given \(f(t)=a+6t\), we have \(f(1)=a+6\), \(f(5)=a+30\), and \(f(9)=a+54\). Their sum is \(3a+90\). Hence, \(3a+90=180\), so \(3a=90\) and \(a=30\). If 36 were used, the sum would be 198, so it is not correct. Exam tip: Evaluate the function at each specified input first, then form the equation using their sum.

Given \(g(t)=b-5t\), we get \(g(2)=b-10\), \(g(6)=b-30\), and \(g(10)=b-50\). Their sum is \((b-10)+(b-30)+(b-50)=3b-90\). Thus, \(3b-90=210\), so \(3b=300\) and \(b=100\). If 105 were used, the sum would be \(225\), not 210. Exam tip: Write each function value separately before combining like terms.

In \(P(t)=120-5t\), the power of \(t\) is 1 and its coefficient is negative, so the value decreases by 5 per unit time. \(120-5t^2\) is not linear. Exam tip: a linear model always has degree 1.

In linear decay, the decrease per equal time interval is constant, so the graph has a fixed negative slope. Check differences between consecutive values; equal percentage decrease indicates exponential decay.

For the models to be equal, set \(36+9t=126-6t\). Moving \(6t\) to the left gives \(15t=90\), so \(t=6\). If \(t=5\) is substituted, the two model values are not equal. Exam tip: to find when two linear models meet, equate their expressions and solve for \(t\).

For equality, set \(52+4t=148-8t\). Moving \(8t\) to the left gives \(12t=96\), so \(t=8\). On checking, \(P(8)=52+32=84\) and \(Q(8)=148-64=84\). At \(t=10\), the two values are not equal. Exam tip: while equating linear expressions, collect all terms containing \(t\) on one side and constants on the other.

Given f(x)=a+12x and f(4)=95, we get a+12(4)=95. Thus, a+48=95, so a=47. Now, f(10)=47+12(10)=47+120=167. Therefore, option C is correct. The value 155 can result from not determining the constant a correctly. Exam tip: first use the given function value to find the unknown constant, then substitute the required value of x.

Given \(g(x)=b-11x\). Substituting \(x=5\), \(g(5)=b-55=103\), so \(b=158\). Now, for \(x=12\), \(g(12)=158-11(12)=158-132=26\). Hence, 26 is correct. An option such as 37 can result from an incorrect subtraction. Exam tip: first find the unknown constant \(b\) from the given function value, then substitute the required value of \(x\).

In \(y=18-2.5x\), the constant term 18 is the initial value and the coefficient \(-2.5\) shows an equal decrease per unit of x. Option C is not linear because it contains \(x^2\). Exam tip: for linear decay, x has power 1 and a negative coefficient.

In \(y=250-7x\), the coefficient of x is \(-7\), so y decreases by 7 for every 1-unit increase in x. \(y=250+7x\) shows growth, while an \(x^2\) relation is not linear. Exam tip: linear decay has a negative coefficient of x.

In linear growth, equal increases in x produce the same change in y each time. For example, when x rises by 1, y-differences may be 5, 5, 5. A constant ratio indicates exponential growth. Exam tip: check first differences first.

\(7-3x\) is a linear polynomial with coefficient \(-3\) of x. Thus, for every increase of 1 in x, its value decreases by 3. In \(x^2-3x+7\), the rate of change is not constant. Exam tip: constant growth or decay requires degree 1.