Question

match each graph with its equation.
$y = x^3$
$y = \sqrt3{x}$
$y = \frac{1}{x^2}$
$y = |x|$
$y = x^2$
$y = \sqrt{x}$
$y = \frac{1}{x}$
$y = x$
graphs:
a. graph with two curves, one on left of y-axis (increasing from (-3,0) to (0,3)) and one on right (decreasing from (0,3) to (3,0))
b. graph of a cubic-like curve, increasing from left (negative x, negative y) through (0,0) to right (positive x, positive y)
c. graph of an absolute value function, v - shaped with vertex at (0,0)
d. graph of a parabola opening upwards with vertex at (0,0) or near (0,0)

Explanation:

Step1: Analyze Graph a

Graph a has two branches, one on the left of the y - axis (for \(x<0\)) and one on the right (for \(x > 0\)), and it is symmetric about the y - axis. The function \(y=\frac{1}{x^{2}}\) has the property that \(f(-x)=\frac{1}{(-x)^{2}}=\frac{1}{x^{2}} = f(x)\), so it is even (symmetric about the y - axis). Also, as \(x\) approaches 0, \(y\) approaches \(+\infty\), and as \(|x|\) approaches \(+\infty\), \(y\) approaches 0, which matches the shape of graph a. So \(y = \frac{1}{x^{2}}\) corresponds to graph a.

Step2: Analyze Graph b

Graph b passes through the origin \((0,0)\) and has a shape that is increasing for \(x>0\) and increasing (with a different slope) for \(x < 0\). The function \(y = \sqrt[3]{x}\) (or \(y=x^{\frac{1}{3}}\)) has the property that \(f(-x)=\sqrt[3]{-x}=-\sqrt[3]{x}=-f(x)\), so it is odd (symmetric about the origin). Also, when \(x = 0\), \(y = 0\), and as \(x\) increases, \(y\) increases (slowly for large \(x\)), and as \(x\) decreases (negative), \(y\) decreases, which matches graph b. So \(y=\sqrt[3]{x}\) corresponds to graph b.

Step3: Analyze Graph c

Graph c is a V - shaped graph with the vertex at the origin \((0,0)\) and symmetric about the y - axis. The function \(y = |x|\) is defined as \(y=

$$\begin{cases}x, & x\geq0\\-x, & x < 0\end{cases}$$

\), which has a V - shape with the vertex at \((0,0)\) and is symmetric about the y - axis. So \(y = |x|\) corresponds to graph c.

Step4: Analyze Graph d

Graph d is a parabola opening upwards with the vertex at the origin \((0,0)\). The function \(y=x^{2}\) is a parabola opening upwards with vertex at \((0,0)\) (since for \(y = ax^{2}+bx + c\), here \(a = 1\), \(b = 0\), \(c = 0\)), and it is even (symmetric about the y - axis). So \(y=x^{2}\) corresponds to graph d.

Step5: Analyze \(y=x^{3}\)

The function \(y=x^{3}\) is odd (\(f(-x)=(-x)^{3}=-x^{3}=-f(x)\)) and passes through the origin. Its graph has a steeper curve than \(y=\sqrt[3]{x}\). If there was another graph (not shown here completely, but from the set), it would match a graph with a more "steep" odd - function shape.

Step6: Analyze \(y = \sqrt{x}\)

The function \(y=\sqrt{x}\) is defined for \(x\geq0\) only. Its graph starts at \((0,0)\) and increases slowly for \(x>0\).

Step7: Analyze \(y=\frac{1}{x}\)

The function \(y=\frac{1}{x}\) is odd (\(f(-x)=\frac{1}{-x}=-\frac{1}{x}=-f(x)\)) and has two branches, one in the first quadrant (\(x>0,y > 0\)) and one in the third quadrant (\(x<0,y < 0\)), symmetric about the origin.

Step8: Analyze \(y=x\)

The function \(y = x\) is a straight line with slope 1, passing through the origin, symmetric about the origin.

Answer:

• \(y=\frac{1}{x^{2}}\) - Graph a
• \(y=\sqrt[3]{x}\) - Graph b
• \(y = |x|\) - Graph c
• \(y=x^{2}\) - Graph d
• \(y=x^{3}\): (would match a graph with a steeper odd - function curve)
• \(y=\sqrt{x}\): (defined for \(x\geq0\), increasing from \((0,0)\))
• \(y=\frac{1}{x}\): (two branches, Q1 and Q3)
• \(y = x\): (straight line, slope 1)