Identifying Functions

1. Find the output, $g$, when the input, $r$, is $3$.

$g=-5 r+13$

2. Find the output, $b$, when the input, $a$, is $6$.

$b=-1-7 a$

3. Find the output, $y$, when the input, $x$, is $5$.

$y=5 x-3$

4. Find the output, $k$, when the input, $t$, is $-7$.

$k=10 t-19$

1. $g = -2$

To find the output,$g$, we need to substitute $3$ into the equation for $r$.

$\begin{aligned} g &=-5 r+13 \\ g &=-5 \cdot 3+13 \\ &=-15+13 \\ &=-2 \end{aligned}$

When the input is $3$, the output is $-2$.

2. $g = -43$

To find the output, $b$, we need to substitute $6$ into the equation for $a$.

$\begin{aligned} b &=-1-7 a \\ b &=-1-7 \cdot 6 \\ &=-1-42 \\ &=-43 \end{aligned}$

When the input is $6$, the output is $-43$.

3. $g = 22$

To find the output, $y$, we need to substitute $5$ into the equation for $x$.

$\begin{aligned} y &=5 x-3 \\ y &=5 \cdot 5-3 \\ &=25-3 \\ &=22 \end{aligned}$

When the input is $5$, the output is $22$.

4. $g = -89$

To find the output, $k$, we need to substitute $-7$ into the equation for $t$.

$\begin{aligned} k &=10 t-19 \\ k &=10 \cdot-7-19 \\ &=-70-19 \\ &=-89 \end{aligned}$

When the input is $-7$, the output is $-89$.

1. Find the output, $y$, when the input, $x$, is $7$.

2. Find the output, $y$, when the input, $x$, is $4$.

3. Find the output, $y$, when the input, $x$, is $2$.

4. Find the output, $y$, when the input, $x$, is $-5$.

1. $y =4$

We should look for the point on the graph whose $x$-coordinate is $7$.

The point on the graph whose $x$-coordinate is $7$ is the point $(7,4)$.

When the input is $7$, the output is $4$.

2. $y=1$

We should look for the point on the graph whose $x$-coordinate is $4$.

The point on the graph whose $x$-coordinate is $4$ is the point $(4,1)$.

When the input is $4$, the output is $1$.

3. $y = -2$

We should look for the point on the graph whose $x$-coordinate is $2$.

The point on the graph whose $x$-coordinate is $2$ is the point $(2,-2)$.

When the input is $2$, the output is $-2$

4. $y=7$

We should look for the point on the graph whose $x$-coordinate is $-5$.

The point on the graph whose $x$-coordinate is $-5$ is the point $(-5,7)$.

When the input is $-5$, the output is $7$.

1. Rearrange the equation so $x$ is the independent variable.

$y+6=5(x-4)$

2. Rearrange the equation so $a$ is the independent variable.

$3 a-7=-4 b+1$

3. Rearrange the equation so $u$ is the independent variable.

$4 u+8 w=-3 u+2 w$

4. Rearrange the equation so $r$ is the independent variable.

$q-10=6(r+1)$

1. $y=5 x-26$

To arrive at a correct equation, we have to solve the equation for $y$.

$\begin{aligned} y+6 &=5(x-4) \\ y &=5(x-4)-6 \\ y &=5 x-20-6 \\ y &=5 x-26 \end{aligned}$

The following equation is rearranged so $x$ is the independent variable:

$y=5 x-26$

2. $b=2-\frac{3}{4} a$

To arrive at a correct equation, we have to solve the equation for $b$.

$\begin{aligned} 3 a-7 &=-4 b+1 \\ 4 b-1 &=7-3 a \\ 4 b &=8-3 a \\ b &=\frac{8}{4}-\frac{3 a}{4} \\ b &=2-\frac{3}{4} a \end{aligned}$

The following equation is rearranged so $a$ is the independent variable:

$b=2-\frac{3}{4} a$

3. $w=-\frac{7}{6} u$

To arrive at a correct equation, we have to solve the equation for $w$.

$\begin{aligned} 4 u+8 w &=-3 u+2 w \\ 6 w &=-7 u \\ w &=\frac{-7 u}{6} \\ w &=-\frac{7}{6} u \end{aligned}$

The following equation is rearranged so $u$ is the independent variable:

$w=-\frac{7}{6} u$

4. $q=6 r+16$

To arrive at a correct equation, we have to solve the equation for $q$.

$\begin{aligned} q-10 &=6(r+1) \\ q &=6(r+1)+10 \\ q &=6 r+6+10 \\ q &=6 r+16 \end{aligned}$

The following equation is rearranged so $r$ is the independent variable:

$q=6 r+16$