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:

\(q=6 r+16\)