Question #b8f13a3a - SAT Suite Question Bank

ID: b8f13a3a

Modded SAT Question Bank by Abdullah Mallik

Function $f$ is defined by $f (x) = - a^{x} + b$ , where $a$ and $b$ are constants. In the xy-plane, the graph of $y = f (x) - 12$ has a y-intercept at $(0, - \frac{75}{7})$ . The product of $a$ and $b$ is $StartFraction 320 Over 7 EndFraction$ . What is the value of $a$ ?

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Correct Answer: 20

Rationale

The correct answer is $20$ . It’s given that $f (x) = - a^{x} + b$ . Substituting $- a^{x} + b$ for $f (x)$ in the equation $y = f (x) - 12$ yields $y = - a^{x} + b - 12$ . It’s given that the y-intercept of the graph of $y = f (x) - 12$ is $(0, - \frac{75}{7})$ . Substituting $0$ for $x$ and $- \frac{75}{7}$ for $y$ in the equation $y = - a^{x} + b - 12$ yields $- \frac{75}{7} = - a^{0} + b - 12$ , which is equivalent to $- \frac{75}{7} = - 1 + b - 12$ , or $- \frac{75}{7} = b - 13$ . Adding $13$ to both sides of this equation yields $\frac{16}{7} = b$ . It’s given that the product of $a$ and $b$ is $StartFraction 320 Over 7 EndFraction$ , or $a b = \frac{320}{7}$ . Substituting $StartFraction 16 Over 7 EndFraction$ for $b$ in this equation yields $(a) (\frac{16}{7}) = \frac{320}{7}$ . Dividing both sides of this equation by $StartFraction 16 Over 7 EndFraction$ yields $a equals 20$ .

Question Difficulty: Hard