Question

画图 $y = 2sqrt3{x^2}$ $y = x$

Explanation:

Step1: Analyze \( y = x \)

The function \( y = x \) is a linear function with a slope of \( 1 \) and a \( y \)-intercept of \( 0 \). It passes through the origin \((0,0)\) and has a \( 45^\circ \) angle with the \( x \)-axis. We can plot points like \((1,1)\), \((-1,-1)\) to draw this line.

Step2: Analyze \( y = 2\sqrt[3]{x^2} \)

First, rewrite the function as \( y = 2x^{\frac{2}{3}} \). The domain of this function is all real numbers (\( x \in \mathbb{R} \)) because we can take the cube root of any real number, and then square it. For \( x = 0 \), \( y = 0 \). For positive \( x \), as \( x \) increases, \( x^{\frac{2}{3}} \) increases (since the exponent \( \frac{2}{3} \) is positive), and multiplying by \( 2 \) scales it. For negative \( x \), \( x^2 \) is positive, so \( y \) is also positive (even function, symmetric about the \( y \)-axis). We can calculate some points: when \( x = 1 \), \( y = 2(1)^{\frac{2}{3}} = 2 \); when \( x = -1 \), \( y = 2(-1)^{\frac{2}{3}} = 2((-1)^2)^{\frac{1}{3}} = 2(1)^{\frac{1}{3}} = 2 \); when \( x = 8 \), \( y = 2(8)^{\frac{2}{3}} = 2(4) = 8 \); when \( x = -8 \), \( y = 2(-8)^{\frac{2}{3}} = 2(4) = 8 \).

Step3: Plot the functions

• For \( y = x \): Draw a straight line through \((0,0)\), \((1,1)\), \((-1,-1)\).
• For \( y = 2\sqrt[3]{x^2} \): Plot the points we calculated (\((0,0)\), \((1,2)\), \((-1,2)\), \((8,8)\), \((-8,8)\)) and sketch the curve, noting its symmetry about the \( y \)-axis and its shape (increasing for \( x > 0 \), decreasing for \( x < 0 \) but \( y \) is always non - negative).

Answer:

To draw the graphs:

1. For \( y = x \): Draw a straight line with slope \( 1 \) passing through the origin.
2. For \( y = 2\sqrt[3]{x^2} \): Sketch a curve symmetric about the \( y \)-axis, passing through \((0,0)\), \((1,2)\), \((-1,2)\), and increasing for \( x>0 \), decreasing for \( x < 0 \) (but \( y\geq0 \) for all \( x \)).