The equation is in slope-intercept form $y=mx+b,$ where $m$ is the slope and $b$ is the $y$-intercept.

For the equation $y=4x - 3,$ we see that the slope is $m=\bbox[4pt,border: 1px solid lightgray]{4}$ and the $y$-intercept is $b=\bbox[4pt,border: 1px solid lightgray]{-3}.$

The equation is in slope-intercept form $y=mx+b,$ where $m$ is the slope and $b$ is the $y$-intercept.

For the equation $y=-9x + 17,$ we see that the slope is $m=-9$ and the $y$-intercept is $b=17.$

Therefore, the point where the given line intersects the $y$-axis is $(0,17).$

The equation is in slope-intercept form $y=mx+b,$ where $m$ is the slope and $b$ is the $y$-intercept.

Note that

$$ y = 3 = 0x + 3. $$

For the equation $y = 0x + 3,$ we see that the slope is $m=0$ and the $y$-intercept is $b=3.$

We will write the equation of the line in the slope-intercept form $y=mx+b.$ We know that the slope is $m=2,$ so we just need to work out $y$-intercept $b.$

From the graph, we see that the line intercepts the $y$-axis at $(0,-3).$ So, the $y$-intercept is $b=-3.$

Substituting the slope $m=2$ and the $y$-intercept $b=-3$ into the slope-intercept form $y=mx+b,$ we obtain the equation of the line:

\[ y=2x-3 \]

We will write the equation of the line in the slope-intercept form $y=mx+b.$ So we need to work out the slope $m$ and the $y$-intercept $b.$

From the graph, we see that the line intercepts the $y$-axis at $(0,1).$ So, the $y$-intercept is $b=1.$

We can calculate the slope of the line by choosing two points on the line and applying the slope formula. The line passes through the points $(0,1)$ and $(1,3),$ so we calculate its slope as follows: \begin{align*} m &= \dfrac{y_2-y_1}{x_2-x_1}\\[5pt] &= \dfrac{3-1}{1-0} \\[5pt] &= \dfrac{2}{1} \\[5pt] &= 2 \end{align*}

Substituting the slope $m=2$ and the $y$-intercept $b=1$ into the slope-intercept form $y=mx+b,$ we obtain the equation of the line:

\[ y=2x+1 \]

We will write the equation of the line in the slope-intercept form $y=mx+b.$ We know that the slope is $m=\dfrac23,$ so we just need to work out $y$-intercept $b.$

From the graph, we see that the line passes through the point $(3,3).$ Substituting this into $y=mx+b$ with $m=\dfrac23$ and solving for $b,$ we get

\begin{align*} y &= \dfrac23x + b \\[5pt] 3 &=\dfrac23(3) + b \\[5pt] 3&= 2 + b \\[5pt] b &= 1. \end{align*}

Therefore, the equation of the line is $y =\bbox[4pt,border: 1px solid lightgray]{\dfrac{2}{3}x + 1}.$

We will write the equation of the line in the slope-intercept form $y=mx+b.$ We know that the slope is $m=-2,$ so we just need to work out $y$-intercept $b.$

From the graph, we see that the line intercepts the $y$-axis at $(0,-1).$ So, the $y$-intercept is $b=-1.$

Substituting the slope $m=-2$ and the $y$-intercept $b=-1$ into the slope-intercept form $y=mx+b,$ we obtain the equation of the line:

\[ y=-2x-1 \]

We will write the equation of the line in the slope-intercept form $y=mx+b.$ So, we need to work out the slope $m$ and the $y$-intercept $b.$

From the graph, we see that the line intercepts the $y$-axis at $(0,5).$ So, the $y$-intercept is $b=5.$

We can calculate the slope of the line by choosing two points on the line and applying the slope formula. The line passes through the points $(0,5)$ and $(5,0),$ so we calculate its slope as follows: \begin{align*} m &= \dfrac{y_2-y_1}{x_2-x_1} \\[5pt] &= \dfrac{0-(5)}{5-0} \\[5pt] &= -1 \end{align*}

Substituting the slope $m=-1$ and the $y$-intercept $b=5$ into the slope-intercept form $y=mx+b,$ we obtain the equation of the line:

\[ y=-x+5 \]

We will write the equation of the line in the slope-intercept form $y=mx+b.$ So, we need to work out the slope $m$ and the $y$-intercept $b.$

First, we calculate the slope $m$ using the given points: \[ \begin{align} m &= \frac{y_2 - y_1}{x_2 - x_1} \\[5pt] &=\frac{26 - 10}{4 - 0}\\[5pt] &=\frac{16}{4}\\[5pt] &=4 \end{align} \]

The coordinates of the $y$-intercept are always $(0,{\color{blue}{b}}).$ We know that the line passes through the point $(0,{\color{blue}{10}}),$ so the $y$-intercept is $b={\color{blue}{10}}.$

Substituting the slope $m=4$ and the $y$-intercept $b=10$ into the slope-intercept formula $y=mx+b,$ we reach

\[ y = 4x + 10. \]

We will write the equation of the line in the slope-intercept form $y=mx+b.$ So, we need to work out the slope $m$ and the $y$-intercept $b.$

First, we calculate the slope $m$ using the given points: \[ \begin{align} m &= \frac{y_2 - y_1}{x_2 - x_1} \\[5pt] &=\frac{31-7}{0-8} \\[5pt] &=\frac{24}{-8} \\[5pt] &=-3 \end{align} \]

The coordinates of the $y$-intercept are always $(0,{\color{blue}{b}}).$ We know that the line passes through the point $(0,{\color{blue}{31}}),$ so the $y$-intercept is $b={\color{blue}{31}}.$

Substituting the slope $m=-3$ and the $y$-intercept $b=31$ into the slope-intercept formula $y=mx+b,$ we reach

\[ y = -3x + 31. \]

We will write the equation of the line in the slope-intercept form $y=mx+b.$ So, we need to work out the slope $m$ and the $y$-intercept $b.$

First, we calculate the slope $m$ using the given points: \begin{align*} m &= \frac{y_2 - y_1}{x_2 - x_1} \\[5pt] &=\frac{-4-(-10)}{0-3} \\[5pt] &=\frac{-4+10}{0-3} \\[5pt] &=\dfrac{6}{-3} \\[5pt] &=-2 \end{align*}

The coordinates of the $y$-intercept are always $(0,{\color{blue}{b}}).$ We know that the line passes through the point $(0,{\color{blue}{-4}}),$ so the $y$-intercept is $b={\color{blue}{-4}}.$

Substituting the slope $m=-2$ and the $y$-intercept $b=-4$ into the slope-intercept formula $y=mx+b,$ we reach \[ y = \bbox[4pt,border: 1px solid lightgray]{-2x-4}. \]