Question

raphael graphed the functions $g(x)=x+2$ and $f(x)=x-1$. how many units below the $y$-intercept of $g(x)$ is the $y$-intercept of $f(x)$?\
graph of two lines $g(x)$ (green) and $f(x)$ (blue) on a coordinate plane with grid. $g(x)$ has a $y$-intercept at $(0,2)$ and $f(x)$ at $(0,-1)$. the $x$-axis ranges from -5 to 5, $y$-axis from -5 to 5.\
\bigcirc -3 units\
\bigcirc -1 units\
\bigcirc 2 units

Explanation:

Step1: Find y - intercept of \(g(x)\)

For a linear function in the form \(y = mx + b\), the y - intercept is the value of \(y\) when \(x = 0\). For \(g(x)=x + 2\), when \(x = 0\), \(g(0)=0 + 2=2\). So the y - intercept of \(g(x)\) is \(2\).

Step2: Find y - intercept of \(f(x)\)

For the function \(f(x)=x - 1\), when \(x = 0\), \(f(0)=0-1=- 1\). So the y - intercept of \(f(x)\) is \(-1\).

Step3: Calculate the distance

To find how many units below the y - intercept of \(g(x)\) is the y - intercept of \(f(x)\), we find the difference between the y - intercept of \(g(x)\) and the y - intercept of \(f(x)\). That is \(2-(-1)=3\)? Wait, no. Wait, we want to find how many units below the y - intercept of \(g(x)\) (which is at \(y = 2\)) is the y - intercept of \(f(x)\) (which is at \(y=-1\)). The number of units between \(y = 2\) and \(y=-1\) is \(2-(-1)=3\)? Wait, no, let's think again. The y - intercept of \(g(x)\) is \(2\) (on the y - axis, the point is \((0,2)\)) and the y - intercept of \(f(x)\) is \(- 1\) (the point is \((0,-1)\)). To find how many units below \((0,2)\) is \((0,-1)\), we calculate \(2-(-1)=3\)? Wait, no, that's not right. Wait, the vertical distance between \(y = 2\) and \(y=-1\) is \(|2-(-1)|=3\)? Wait, but the options don't have 3. Wait, maybe I made a mistake. Wait, the function \(g(x)=x + 2\), when \(x = 0\), \(y = 2\). The function \(f(x)=x-1\), when \(x = 0\), \(y=-1\). The number of units from \(y = 2\) to \(y=-1\) moving down: from \(y = 2\) to \(y = 0\) is 2 units, and from \(y = 0\) to \(y=-1\) is 1 unit, so total \(2 + 1=3\)? But the options given are \(-3\) units, \(-1\) units, \(-1\) units, 2 units. Wait, maybe the question is asking for the difference in the y - values. Wait, the y - intercept of \(g(x)\) is \(2\), y - intercept of \(f(x)\) is \(-1\). The difference is \(-1-2=-3\). But the question says "how many units below", so the number of units is \(|2-(-1)| = 3\), but the options have \(-3\) units? Wait, maybe the question is considering the direction. If we take the y - intercept of \(f(x)\) minus the y - intercept of \(g(x)\), we get \(-1 - 2=-3\). So the y - intercept of \(f(x)\) is 3 units below the y - intercept of \(g(x)\), which is represented as \(-3\) units (since it's below, the value is negative in terms of the difference). Wait, but the options: let's check the options again. The options are \(-3\) units, \(-1\) units, \(-1\) units, 2 units. Wait, maybe I made a mistake in the functions. Wait, \(g(x)=x + 2\), y - intercept at \(y = 2\); \(f(x)=x-1\), y - intercept at \(y=-1\). The vertical distance between \(y = 2\) and \(y=-1\) is \(2-(-1)=3\), but if we are asked how many units below, the answer is 3, but since the options have \(-3\) units (maybe indicating the direction, as moving down is negative), so the answer is \(-3\) units? Wait, no, the options: let's check the graph. From the graph, \(g(x)\) crosses the y - axis at \(y = 2\), \(f(x)\) crosses at \(y=-1\). The number of units from \(y = 2\) to \(y=-1\) is 3 units down. So the answer should be \(-3\) units (if we consider the displacement as \(f(x)\)'s y - intercept minus \(g(x)\)'s y - intercept: \(-1-2=-3\)).

Answer:

\(-3\) units