QUESTION IMAGE
Question
which line has a slope of \\(\frac{1}{2}\\) and a \\(y\\)-intercept of 2?
Step1: Recall the slope - intercept form
The slope - intercept form of a line is $y = mx + b$, where $m$ is the slope and $b$ is the $y$-intercept. We are given that $m=\frac{1}{2}$ and $b = 2$, so the equation of the line should be $y=\frac{1}{2}x + 2$.
Step2: Analyze the $y$-intercept
The $y$-intercept is the point where the line crosses the $y$-axis, so when $x = 0$, $y=b$. For our line, when $x = 0$, $y=2$. So the line should cross the $y$-axis at $(0,2)$.
Step3: Analyze the slope
The slope $m=\frac{1}{2}$ means that for every 2 units we move to the right (increase in $x$ by 2), we move up 1 unit (increase in $y$ by 1). Let's check the graphs:
- For the first graph: The $y$-intercept is - 2 (since it crosses the $y$-axis at $(0,-2)$), so it's not our line.
- For the second graph: The $y$-intercept is - 4 (crosses the $y$-axis at $(0, - 4)$), so it's not our line.
- For the third graph: The $y$-intercept is 2 (crosses the $y$-axis at $(0,2)$). Let's check the slope. From $(0,2)$, if we move 2 units to the right (to $x = 2$), we move up 1 unit (to $y=3$). The slope between $(0,2)$ and $(2,3)$ is $\frac{3 - 2}{2-0}=\frac{1}{2}$, which matches our slope.
- For the fourth graph: The slope is steeper. Let's check the slope. From $(0,2)$, if we move 1 unit to the right (to $x = 1$), we move up 2 units (to $y = 4$), so the slope is $\frac{4 - 2}{1-0}=2$, which is not $\frac{1}{2}$.
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The line in the third graph (the one with the circle in the bottom - left of the four - graph set, crossing the $y$-axis at $(0,2)$ and having a slope of $\frac{1}{2}$)