Linear growth and decay

In linear growth, the change over each equal time interval is constant, so its model has the form y = mt + c. Growth by a constant factor is exponential, not linear. Exam tip: look for a constant difference to identify linear growth.

In linear decay, the change over each equal time interval is constant, so successive differences are equal. A fixed percentage decrease is exponential decay. Exam tip: check consecutive differences.

For the models to be equal, set \(25+6t=70-3t\). Moving \(3t\) to the left gives \(9t=45\), so \(t=5\). At \(t=4\), the two model values are not equal, making it a close but incorrect option. Exam tip: equate the two expressions first, then collect all variable terms on one side.

For equality, set \(P(t)=Q(t)\): \(40+4t=100-6t\). Adding \(6t\) to both sides gives \(40+10t=100\), so \(10t=60\) and \(t=6\). Hence, option C is correct. At \(t=5\), \(P(t)=60\) and \(Q(t)=70\), so they are not equal. Exam tip: equate the two expressions first, then collect all terms containing \(t\) on one side.

Given \(f(x)=a+7x\) and \(f(4)=53\), we get \(a+7(4)=53\). Thus, \(a+28=53\), so \(a=25\). Now \(f(10)=25+7(10)=25+70=95\). Therefore, 95 is the correct option. Getting 88 would require using an incorrect value of the constant \(a\). Exam tip: First use the given function value to find the unknown constant, then substitute the required input.

Given \(g(x)=b-9x\). Substituting \(x=5\), \(g(5)=b-45=72\), so \(b=117\). Now, substituting \(x=11\), \(g(11)=117-9(11)=117-99=18\). Hence, 18 is the correct option. The nearby distractor 27 would result from incorrectly using \(117-90\) instead of calculating \(9\times11\). Exam tip: first find the unknown constant using the given function value, then substitute the required input.

\(b\) is the coefficient of \(x\), so it gives the change in \(y\) for every 1-unit increase in \(x\). If \(b>0\), \(y\) grows; \(b<0\) indicates decay. Exam tip: check the sign of \(b\), not the constant \(a\).

In \(y=48-6x\), the coefficient of \(x\) is \(-6\), so \(y\) falls by 6 units for every 1-unit increase in \(x\). \(y=48+6x\) shows linear growth. Exam tip: in \(y=a+bx\), \(b<0\) indicates linear decay.

For \(P(x)=a+bx\), increasing \(x\) by 1 changes the value by \(\Delta P=b\). Thus, \(b<0\) gives a decrease at every step. The constant \(a\) only gives the initial value. Exam tip: check the sign of \(b\).

In \(7-2x\), the coefficient of x is \(-2\), so the value falls by the same amount for every unit increase in x. \(2x-7\) shows growth because its coefficient is positive. Exam tip: decay has a negative x-coefficient.

Given \(C(t)=18+kt\), we have \(C(5)=18+5k\) and \(C(2)=18+2k\). Therefore, \(C(5)-C(2)=(18+5k)-(18+2k)=3k\). Since this difference is 21, \(3k=21\), so \(k=7\). Option 21 is the given difference, not the value of \(k\). Exam tip: when subtracting two values of a linear expression, the constant term cancels out.

Given \(D(t)=120-kt\), we have \(D(3)=120-3k\) and \(D(9)=120-9k\). Hence, \(D(3)-D(9)=(120-3k)-(120-9k)=6k\). Since \(6k=42\), \(k=7\). If \(k=6\), the difference would be only \(36\), not 42. Exam tip: when subtracting function values, cancel the common constant terms first.

Given \(A(x)=12+5x\), we have \(A(x+4)=12+5(x+4)\) and \(A(x-1)=12+5(x-1)\). Hence, \(A(x+4)-A(x-1)=5[(x+4)-(x-1)]=5\times5=25\). Choosing \(20\) would count only an increase of \(4\), whereas the actual difference between the two inputs is \(5\). Exam tip: for a linear function \(mx+c\), the constant term \(c\) cancels when taking a difference.

In a linear polynomial, \(b\) is the rate of change with respect to \(x\). If \(b<0\), \(P(x)\) decreases for every increase of 1 in \(x\), so it represents decay. \(b=0\) represents a constant quantity. Exam tip: check the sign of the coefficient of \(x\).

Given P(t)=p+6t, substitute P(3)=50 first: p+18=50, so p=32. Now use P(s)=80: 32+6s=80, giving 6s=48 and s=8. If s were 7, then P(7)=32+42=74, not 80. Exam tip: find the unknown constant p from one condition before using the second condition.

Given \(R(t)=r-5t\), substitute \(t=4\) and \(R(4)=90\): \(r-5(4)=90\), so \(r=110\). Next, \(R(s)=55\) gives \(110-5s=55\). Hence \(5s=55\) and \(s=11\). Therefore, \(11\) is correct. If \(s=10\), the model gives 60, not 55. Exam tip: first use the known time-value pair to find the constant \(r\), then use the second value to find the unknown time.

An increase of 4 per step means the linear expression must have the form \(y=a+4x\). Using \(y(3)=29\), we get \(a+4(3)=29\), so \(a=17\). Hence, \(y=17+4x\) is correct. In option B, \(y(3)=41\), while option C represents a decrease rather than an increase. Exam tip: Substitute the given value of \(x\) into each option to check the condition quickly.

A decrease of 6 per step means that the coefficient of x must be \(-6\). Let \(y=a-6x\). Using \(y(5)=42\), we get \(a-6(5)=42\), so \(a=72\). Hence, \(y=72-6x\) is correct. Option A has the correct slope, but it gives \(y(5)=12\). Exam tip: first check the slope for growth or decay, then substitute the given point to verify the constant term.

The required condition is \(M(t)=N(t)+10\). So, \(35+5t=65-3t+10\), or \(35+5t=75-3t\). This gives \(8t=40\), hence \(t=5\). At \(t=4\), the difference is only \(2\), so it is not correct. Exam tip: translate “10 more than” as \(M=N+10\).

“12 more” means \(A(t)-B(t)=12\). Therefore, \((90-4t)-(30+2t)=12\), so \(60-6t=12\). Hence, \(6t=48\) and \(t=8\). At \(t=10\), the two values become equal, so it is a close but incorrect option. Exam tip: In “more than” questions, first write the difference between the two quantities.