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Algebra / Linear functions Difficulty: Hard

$x$ $x$	$f (x)$ $f (x)$
$- 4$ $negative 4$	$0$ $0$
$- \frac{19}{5}$ $- \frac{19}{5}$	$1$ $1$
$- \frac{18}{5}$ $- \frac{18}{5}$	$2$ $2$

For the linear function $f$ , the table shows three values of $x$ and their corresponding values of $f (x)$ . If $h (x) = f (x) - 13$ , which equation defines $h$ ?

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Explanation

Choice B is correct. An equation that defines a linear function $f$ can be written in the form $f (x) = m x + b$ , where $m$ and $b$ are constants. It's given in the table that when $x = - 4$ , $f left parenthesis x right parenthesis equals 0$ . Substituting $negative 4$ for $x$ and $0$ for $f (x)$ in the equation $f (x) = m x + b$ yields $0 = m (- 4) + b$ , or $0 = - 4 m + b$ . Adding $4 m$ to both sides of this equation yields $4 m equals b$ . Substituting $4 m$ for $b$ in the equation $f (x) = m x + b$ yields $f (x) = m x + 4 m$ . It's also given in the table that when $x = - \frac{19}{5}$ , $f left parenthesis x right parenthesis equals 1$ . Substituting $- \frac{19}{5}$ for $x$ and $1$ for $f (x)$ in the equation $f (x) = m x + 4 m$ yields $1 = m (- \frac{19}{5}) + 4 m$ , or $1 equals one fifth m$ . Multiplying both sides of this equation by $5$ yields $m equals 5$ . Substituting $5$ for $m$ in the equation $f (x) = m x + 4 m$ yields $f (x) = 5 x + 4 (5)$ , or $f (x) = 5 x + 20$ . If $h (x) = f (x) - 13$ , substituting $5 x plus 20$ for $f (x)$ in this equation yields $h (x) = (5 x + 20) - 13$ , or $h (x) = 5 x + 7$ .

Choice A is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect. This is an equation that defines the linear function $f$ , not $h$ .