Find the equation whose graph is shown below. Write your answer in standard form.
(Standard form is , where is positive, and , , and are integers with greatest common divisor 1.)
From the graph, we can see that the line passes through the points $(0,-3)$ and $(-3,3)$.
Standard form is $y = mx+b$, where $b$ is a constant and $m$ is the slope of the line.
The slope of the line is given by $\frac{y_1-y_2}{x_1-x_2}$. Thus, the slope of this line is $\frac{-3-3}{0-(-3)} = -2$
Plugging a point into our new equation of $y = -2x + b$ gives $-3 = 0 + (-3)$, meaning $b = -3$.
Thus, our final equation is $\boxed{y = -2x-3}$
From the graph, we can see that the line passes through the points $(0,-3)$ and $(-3,3)$.
Standard form is $y = mx+b$, where $b$ is a constant and $m$ is the slope of the line.
The slope of the line is given by $\frac{y_1-y_2}{x_1-x_2}$. Thus, the slope of this line is $\frac{-3-3}{0-(-3)} = -2$
Plugging a point into our new equation of $y = -2x + b$ gives $-3 = 0 + (-3)$, meaning $b = -3$.
Thus, our final equation is $\boxed{y = -2x-3}$