Question

question
write the equation of the line in fully simplified slope - intercept form.

Explanation:

Step1: Identify two points on the line

From the graph, we can see that the line passes through the points \((0, -1)\) (the y - intercept) and \((2, 4)\).

Step2: Calculate the slope (\(m\))

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let \((x_1,y_1)=(0, - 1)\) and \((x_2,y_2)=(2,4)\). Then \(m=\frac{4-(-1)}{2 - 0}=\frac{4 + 1}{2}=\frac{5}{2}\)? Wait, no, wait. Wait, let's check another point. Wait, when \(x = 0\), \(y=-1\); when \(x = 2\), \(y = 4\)? Wait, no, maybe I made a mistake. Wait, looking at the graph again, when \(x = 0\), the line crosses the y - axis at \(y=-1\)? Wait, no, wait the line passes through \((0, - 1)\) and \((1,2)\)? Wait, no, let's take two clear points. Let's take \((0,-1)\) and \((2,4)\) no, wait, when \(x = 2\), the y - value is 4? Wait, no, maybe I misread. Wait, let's take \((0,-1)\) and \((1,2)\). Wait, slope \(m=\frac{2-(-1)}{1 - 0}=\frac{3}{1}=3\). Wait, let's check with another pair. Let's take \((-1,-4)\) and \((0,-1)\). Then \(m=\frac{-1-(-4)}{0-(-1)}=\frac{-1 + 4}{1}=3\). Ah, right. So the slope \(m = 3\).

Step3: Determine the y - intercept (\(b\))

The slope - intercept form is \(y=mx + b\), where \(b\) is the y - intercept. From the graph, when \(x = 0\), \(y=-1\), so \(b=-1\).

Step4: Write the equation

Substitute \(m = 3\) and \(b=-1\) into the slope - intercept form \(y=mx + b\). We get \(y = 3x-1\). Wait, wait, let's verify with a point. Take \(x = 1\), \(y=3(1)-1=2\). Looking at the graph, when \(x = 1\), \(y = 2\), which matches. Take \(x = 2\), \(y=3(2)-1 = 5\)? Wait, no, in the graph, when \(x = 2\), the point is at \(y = 4\)? Wait, I must have misread the points. Wait, let's re - examine the graph. The line passes through \((0,-1)\), \((1,2)\), \((2,5)\)? No, wait the original graph: when \(x = 0\), \(y=-1\); \(x = 1\), \(y = 2\); \(x = 2\), \(y = 5\)? Wait, no, the user's graph: looking at the grid, each square is 1 unit. So when \(x = 0\), the line is at \(y=-1\); \(x = 1\), \(y = 2\); \(x = 2\), \(y = 5\)? Wait, no, maybe my initial point selection was wrong. Wait, let's take \((0,-1)\) and \((1,2)\): slope is \(\frac{2-(-1)}{1-0}=3\). Then \(y=3x + b\). Plug in \((0,-1)\): \(-1=3(0)+b\), so \(b=-1\). Then \(y = 3x-1\). Let's check \(x = 2\): \(y=3(2)-1 = 5\). But in the graph, when \(x = 2\), the point is at \(y = 4\)? Wait, no, maybe the graph has a different scale. Wait, maybe I made a mistake in the points. Wait, let's look at the graph again. The line passes through \((0,-1)\), \((1,2)\), \((2,5)\)? No, the blue line: when \(x = 0\), it's at \(y=-1\); \(x = 1\), \(y = 2\); \(x = 2\), \(y = 5\); \(x = 3\), \(y = 8\); \(x = 4\), \(y = 11\). Yes, that makes sense. So the slope is 3, y - intercept is - 1. So the equation is \(y=3x - 1\). Wait, but let's check with \((-1,-4)\): \(y=3(-1)-1=-3 - 1=-4\), which matches. So that's correct.

Answer:

\(y = 3x-1\)