which is the graph of $y = \frac{2}{x} + 2$?

Question

Accepted Answer

# Explanation: ## Step1: Analyze the parent function The given function is $ y=\frac{2}{x}+2 $. The parent function is $ y = \frac{2}{x} $, which is a hyperbola with vertical asymptote $ x = 0 $ and horizontal asymptote $ y=0 $. ## Step2: Determine the transformations For the function $ y=\frac{2}{x}+2 $, we have a vertical shift of the parent function $ y = \frac{2}{x} $ by 2 units upward. So the horizontal asymptote of $ y=\frac{2}{x}+2 $ will be $ y = 2 $ (since we shift the horizontal asymptote $ y = 0 $ of the parent function up by 2 units), and the vertical asymptote remains $ x=0 $ (the y - axis). ## Step3: Analyze the quadrants For the parent function $ y=\frac{2}{x} $, when $ x>0 $, $ y>0 $ (so the graph is in the first quadrant), and when $ x < 0 $, $ y<0 $ (so the graph is in the third quadrant). After the vertical shift $ y=\frac{2}{x}+2 $, when $ x>0 $, $ y=\frac{2}{x}+2>0 + 2=2>0 $, so the part of the graph for $ x>0 $ is still in the first quadrant (above $ y = 2 $ when $ x$ is small positive). When $ x<0 $, $ y=\frac{2}{x}+2$. Let's take $ x=- 1$, then $ y=\frac{2}{-1}+2=-2 + 2=0$; when $ x=-2$, $ y=\frac{2}{-2}+2=-1 + 2 = 1>0$. So for $ x<0 $, the graph of $ y=\frac{2}{x}+2$ is in the second quadrant (since when $ x<0$, $ y=\frac{2}{x}+2$ can be positive: for example, when $ x=-1$, $ y = 0$; when $ x=-0.5$, $ y=\frac{2}{-0.5}+2=-4 + 2=-2<0$? Wait, no, let's recalculate. If $ x=-0.5$, $ y=\frac{2}{-0.5}+2=-4 + 2=-2$. Wait, maybe my earlier thought was wrong. Wait, the parent function $ y = \frac{2}{x}$ for $ x<0$ has $ y<0 $. After adding 2, $ y=\frac{2}{x}+2$. Let's find when $ y = 0$: $ \frac{2}{x}+2=0\Rightarrow\frac{2}{x}=-2\Rightarrow x=-1$. So when $ x < - 1$, $ \frac{2}{x}>-2$ (because if $ x=-2$, $ \frac{2}{-2}=-1>-2$), so $ y=\frac{2}{x}+2>-2 + 2=0$. When $ - 10$ (second quadrant), for $ - 10$, the graph is in the first quadrant (above $ y = 2$ when $ x$ is small), and for $ x<0$, the graph is in the second quadrant (since when $ x<0$, the graph is above the x - axis for $ x < - 1$ and crosses the x - axis at $ x=-1$ and goes below for $ - 10$ a curve in the first quadrant (starting near $ y$ - axis, going down towards $ x$ - axis as $ x$ increases) and for $ x<0$ a curve in the second quadrant (starting near $ y$ - axis, going down towards $ x$ - axis as $ x$ becomes more negative). Wait, the key is the horizontal asymptote. The function $ y=\frac{2}{x}+2$ has a horizontal asymptote at $ y = 2$. So as $ x\to\pm\infty$, $ y\to2$. So we need to find the graph where as $ x$ becomes very large positive or very large negative, the graph approaches $ y = 2$. Looking at the four graphs: - The first graph: As $ x\to\infty$, the curve approaches $ y = 0$? No, wait, no. Wait, let's re - evaluate. The first graph (leftmost) has for $ x>0$ a curve that starts near the $ y$ - axis (high $ y$ - value) and decreases towards the $ x$ - axis as $ x$ increases. For $ x<0$, a curve that starts near the $ y$ - axis (low $ y$ - value, below $ y=-2$) and increases towards the $ x$ - axis as $ x$ becomes more negative. No, that's not our function. - The second graph: For $ x>0$, the curve is in the first quadrant, and as $ x\to\infty$, it approaches $ y = 0$? No. - The third graph: For $ x>0$, the curve is in the first quadrant, and as $ x\to\infty$, it approaches $ y = 2$ (since it's getting closer to the line $ y = 2$). For $ x<0$, the curve is in the second quadrant (above the x - axis for $ x < - 1$) and as $ x\to-\infty$, it approaches $ y = 2$. - The fourth graph: For $ x>0$, the curve is in the fourth quadrant (below $ x$ - axis), which is wrong because for $ x>0$, $ y=\frac{2}{x}+2>0$. So the correct graph is the third one (assuming the order is left to right: first, second, third, fourth). Wait, maybe the first graph (leftmost) is the correct one? Wait, no, let's take a point. Let's find a point on the function $ y=\frac{2}{x}+2$. When $ x = 1$, $ y=\frac{2}{1}+2=4$. When $ x = 2$, $ y=\frac{2}{2}+2=1 + 2=3$. When $ x=-1$, $ y=\frac{2}{-1}+2=0$. When $ x=-2$, $ y=\frac{2}{-2}+2=-1 + 2=1$. Now let's check the first graph (leftmost): - For $ x = 1$, the $ y$ - value is around 4 (matches $ y = 4$ when $ x = 1$). - For $ x = 2$, the $ y$ - value is around 3 (matches $ y = 3$ when $ x = 2$). - For $ x=-1$, the $ y$ - value is 0 (matches our calculation). - For $ x=-2$, the $ y$ - value is around 1 (matches our calculation). - The horizontal asymptote: As $ x\to\pm\infty$, $ y\to2$. In the first graph, as $ x\to\infty$, the curve is approaching $ y = 2$ (since it's getting closer to the line $ y = 2$ as $ x$ increases), and as $ x\to-\infty$, the curve is approaching $ y = 2$ (since it's getting closer to the line $ y = 2$ as $ x$ becomes more negative). The other graphs: - The second graph: When $ x = 1$, $ y$ is around 4? No, the second graph's curve for $ x>0$ is higher up, but when $ x=-1$, the $ y$ - value is negative, which doesn't match our function (at $ x=-1$, $ y = 0$). - The third graph: When $ x=-1$, the $ y$ - value is negative, which is wrong. - The fourth graph: When $ x = 1$, the $ y$ - value is negative, which is wrong. So the correct graph is the first one (leftmost) among the four. # Answer: The left - most graph (the first graph in the given set of four graphs) is the graph of $ y=\frac{2}{x}+2$.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Analyze the parent function

Step2: Determine the transformations

Step3: Analyze the quadrants

Answer: