Linear functions with an unknown constant Practice Questions

Warm-upmedium

In the rule $g(x) = 8x + c$ , the value of $c$ is a constant. Given that $g(3) = 10$ , find $g(5)$ .

Substitute $x = 3$ : $8(3) + c = 10$ , so $24 + c = 10$ and $c = -14$ . The rule is $g(x) = 8x - 14$ , so $g(5) = 8(5) - 14 = 40 - 14 = 26$ .

Mid-gamemedium

Let $p(x) = kx - 6$ for some constant $k$ . If $p(2) = 10$ , what is the value of $p(5)$ ?

Substitute $x = 2$ : $2k - 6 = 10$ , so $2k = 16$ and $k = 8$ . The rule is $p(x) = 8x - 6$ , so $p(5) = 8(5) - 6 = 40 - 6 = 34$ .

Mid-gamemedium

Consider the linear function $t$ given by $t(x) = nx + 150$ , where $n$ is a constant. Knowing that $t(10) = 80$ , determine $t(20)$ .

Substitute $x = 10$ : $10n + 150 = 80$ , so $10n = -70$ and $n = -7$ . The rule is $t(x) = -7x + 150$ , so $t(20) = -7(20) + 150 = -140 + 150 = 10$ .

Mid-gamemedium

For the function $v(x) = 6x + k$ with constant $k$ , it is known that $v(7) = 30$ . What number does $v(4)$ equal?

Substitute $x = 7$ : $6(7) + k = 30$ , so $42 + k = 30$ and $k = -12$ . The rule is $v(x) = 6x - 12$ , so $v(4) = 6(4) - 12 = 24 - 12 = 12$ .

Mid-gamemedium

Here $c$ is a constant in the rule $w(x) = cx - 15$ . Because $w(4) = 5$ , what must $w(9)$ be?

Substitute $x = 4$ : $4c - 15 = 5$ , so $4c = 20$ and $c = 5$ . The rule is $w(x) = 5x - 15$ , so $w(9) = 5(9) - 15 = 45 - 15 = 30$ .

Mid-gamemedium

The linear function $h$ is defined by $h(x) = mx + 45$ , where $m$ is a constant. Given that $h(9) = 0$ , evaluate $h(4)$ .

Substitute $x = 9$ : $9m + 45 = 0$ , so $9m = -45$ and $m = -5$ . The rule is $h(x) = -5x + 45$ , so $h(4) = -5(4) + 45 = -20 + 45 = 25$ .

Mid-gamehard

The rule $r(x) = -4x + d$ describes a linear function, and $d$ is a constant. If $r(3) = 5$ , what is $r(-2)\textsf{?}$

Substitute $x = 3$ : $-4(3) + d = 5$ , so $-12 + d = 5$ and $d = 17$ . The rule is $r(x) = -4x + 17$ , so $r(-2) = -4(-2) + 17 = 8 + 17 = 25$ .

Buzzer beaterhard

Suppose $z(x) = -2x + b$ , where $b$ is a constant. If $z(8) = 3$ , what is the value of $z(1)\textsf{?}$

Substitute $x = 8$ : $-2(8) + b = 3$ , so $-16 + b = 3$ and $b = 19$ . The rule is $z(x) = -2x + 19$ , so $z(1) = -2(1) + 19 = -2 + 19 = 17$ .

Linear functions with an unknown constant. Game on.