sat suite question viewer

Algebra / Linear functions Difficulty: Easy

$x$ $x$	$f (x)$ $f (x)$
$0$ $0$	$29$ $29$
$1$ $1$	$32$ $32$
$2$ $2$	$35$ $35$

For the linear function $f$ , the table shows three values of $x$ and their corresponding values of $f (x)$ . Which equation defines $f (x)$ ?

$f (x) = 3 x + 29$

$f (x) = 29 x + 32$

$f (x) = 35 x + 29$

$f (x) = 32 x + 35$

Back question 36 of 152 Next

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152

Explanation

Choice A 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 $x equals 0$ , $f left parenthesis x right parenthesis equals 29$ . Substituting $0$ for $x$ and $29$ for $f (x)$ in the equation $f (x) = m x + b$ yields $29 = m (0) + b$ , or $29 equals b$ . Substituting $29$ for $b$ in the equation $f (x) = m x + b$ yields $f (x) = m x + 29$ . It's also given in the table that when $x equals 1$ , $f left parenthesis x right parenthesis equals 32$ . Substituting $1$ for $x$ and $32$ for $f (x)$ in the equation $f (x) = m x + 29$ yields $32 = m (1) + 29$ , or $32 = m + 29$ . Subtracting $29$ from both sides of this equation yields $3 equals m$ . Substituting $3$ for $m$ in the equation $f (x) = m x + 29$ yields $f (x) = 3 x + 29$ .

Choice B 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 and may result from conceptual or calculation errors.