match the functions: $y = |x|$, $y = \frac{1}{x}$, $y = \frac{1}{x^2}$, $y = x^2$, $y = x^3$, $y = x$, $y = \sqrt3{x}$, $y = \sqrt{x}$ to the graphs labeled a, b, c, d (and other possible graphs not fully shown here).

Question

Accepted Answer

To solve this problem, we analyze each graph and match it with the corresponding function:

Part a:

The graph in part a is a V - shaped graph with the vertex at the origin $(0,0)$. The function $y = |x|$ (the absolute - value function) has a graph that is a V - shaped graph opening upwards with the vertex at the origin. For $x\geq0$, $y=x$ (a line with a slope of 1), and for $x < 0$, $y=-x$ (a line with a slope of - 1), which matches the graph in part a.

Part b:

The graph in part b has a point - symmetric shape about the origin. The function $y=x^{3}$ is an odd function, and its graph is symmetric about the origin. When $x = 0$, $y = 0$. For $x>0$, $y=x^{3}$ is an increasing function, and for $x < 0$, $y=x^{3}$ is also an increasing function (since if \(x_1

Part c:

The graph in part c is a parabola opening upwards with the vertex at the origin $(0,0)$. The function $y = x^{2}$ is a quadratic function, and its graph is a parabola opening upwards with the vertex at the origin. For any real number $x$, $y=x^{2}\geq0$, and the graph is symmetric about the $y$ - axis, which matches the graph in part c.

Part d:

The graph in part d is a hyperbola in the first and third quadrants (for $y=\frac{1}{x}$) or in the first and second quadrants (for $y = \frac{1}{x^{2}}$). The function $y=\frac{1}{x^{2}}$ has a graph that is symmetric about the $y$ - axis. When $x>0$, as $x$ increases, $y=\frac{1}{x^{2}}$ decreases towards 0, and when $x < 0$, as $|x|$ increases, $y=\frac{1}{x^{2}}$ also decreases towards 0. The graph in part d (assuming it's the graph of $y=\frac{1}{x^{2}}$) or if it's the graph of $y = \frac{1}{x}$, the graph of $y=\frac{1}{x}$ has two branches, one in the first quadrant ($x>0,y>0$) and one in the third quadrant ($x < 0,y < 0$). But looking at the options and the shape, if we consider the graph in part d (the last one), the function $y=\frac{1}{x^{2}}$ has a graph that is above the $x$ - axis (since $x^{2}>0$ for $x
eq0$) and symmetric about the $y$ - axis. However, if we consider the graph of $y=\frac{1}{x}$, it has two branches in the first and third quadrants. But from the given options and the shape of the graph in part d (the one with the curve in the first and second quadrants is not $y = \frac{1}{x}$, the graph of $y=\frac{1}{x^{2}}$ is in the first and second quadrants (since $x^{2}>0$ for all $x
eq0$, so $y=\frac{1}{x^{2}}>0$ for all $x
eq0$). The graph in part d (the last one) with the curve in the first quadrant (and for $x<0$, since $x^{2}$ is positive, $y=\frac{1}{x^{2}}$ is positive) matches $y=\frac{1}{x^{2}}$ or $y=\frac{1}{x}$. But if we re - examine, the graph of $y=\frac{1}{x}$ has a vertical asymptote at $x = 0$ and horizontal asymptote at $y = 0$, with two branches (first and third quadrants). The graph of $y=\frac{1}{x^{2}}$ has a vertical asymptote at $x = 0$ and horizontal asymptote at $y = 0$, with two branches (first and second quadrants).

Final Matching:

a: $y = |x|$
b: $y=x^{3}$
c: $y = x^{2}$
d: $y=\frac{1}{x^{2}}$ (or $y=\frac{1}{x}$ depending on the exact shape, but from the given options and the most likely match, for the graph in part d, $y=\frac{1}{x^{2}}$ is a better match as it is symmetric about the $y$ - axis and all $y$ - values are positive for \(x

eq0\))

If we assume the questio…

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Part a:

Part b:

Part c:

Part d:

Final Matching:

Answer: