Exponential Functions: math test solutions, Grade 11 – download the PDF

1 Classify and evaluate / 3 pts

1. \( b = 1.04 > 1 \): growth of 4% (0.5 pt). \( b = 0.95 \): decay of 5% (0.5 pt).
2. \( f(0) = 4 \) (0.5 pt). \( f(2) = 4 \cdot \dfrac{9}{4} = 9 \) (0.75 pt). \( f(-1) = 4 \cdot \dfrac{2}{3} = \dfrac{8}{3} \) (0.75 pt).

2 Transformations and graph / 4 pts

1. Shift 1 unit left and 3 units down (1 pt).
2. Asymptote \( y = -3 \) (0.5 pt); range \( y > -3 \) (0.5 pt).
3. \( g(0) = 2^{1} - 3 = -1 \), so \( (0, -1) \) (1 pt).
4. The values are \( -2.75, -2.5, -2, -1, 1 \) (0.5 pt); the curve is below.

The curve rises to the right and approaches the dashed asymptote on the left (0.5 pt).

3 Equation from two points / 3 pts

Divide the equations: \( \dfrac{ab^{3}}{ab} = b^{2} = \dfrac{108}{12} = 9 \) (1 pt).

With \( b > 0 \), \( b = 3 \) (0.5 pt). Then \( 3a = 12 \), so \( a = 4 \) (0.5 pt). Formula \( f(x) = 4 \cdot 3^{x} \) (0.5 pt).

\( f(4) = 4 \cdot 81 = 324 \) (0.5 pt).

4 Compound interest (calculator allowed) / 4 pts

(a) \( A = 3500\left(1 + \dfrac{0.032}{4}\right)^{20} = 3500(1.008)^{20} \approx 4104.67 \) dollars (2 pts).

(b) \( A = 3500\,e^{0.032 \cdot 5} = 3500\,e^{0.16} \approx 4107.29 \) dollars (1 pt).

(c) Using the unrounded balances, \( 4107.288 - 4104.674 pprox 2.61 \) dollars more (1 pt).

5 A laptop loses value / 3 pts

(a) \( V(t) = 1200(0.75)^{t} \) (1 pt).

(b) \( V(2) = 1200 \cdot 0.5625 = 675 \) dollars (1 pt).

(c) \( V(3) = 1200 \cdot 0.421875 = 506.25 < 600 \) while \( V(2) = 675 > 600 \): year 3 (1 pt).

6 Linking e and money / 3 pts

(a) \( 1.01^{100} \approx 2.7048 \) (0.5 pt), slightly under \( e \approx 2.7183 \), a gap of about 0.0135 (0.5 pt).

(b) \( A = 600\,e^{0.05 \cdot 8} = 600\,e^{0.4} \approx 600 \cdot 1.49182 \approx 895.09 \) dollars (2 pts: 1 for the formula, 1 for the value).