Graph natural log of 4x^2

Problem

y=ln(4*x^2)

Solution

1. Identify the domain of the function. Since the argument of a natural logarithm must be positive, we require 4*x^2>0 This is true for all x except x=0

2. Determine symmetry by checking if the function is even or odd. Since ƒ*(−x)=ln(4*(−x)^2)=ln(4*x^2) the function is even and symmetric about the yaxis.

3. Find the intercepts by setting y=0 Solving 0=ln(4*x^2) gives e^0=4*x^2 which simplifies to 1=4*x^2 so x=±1/2

4. Identify vertical asymptotes by checking where the argument approaches zero. As x→0 4*x^2→0^+ so y→−∞ There is a vertical asymptote at x=0

5. Analyze the behavior as x→±∞ As |x| increases, 4*x^2 increases, so y→∞

6. Determine the derivative to find the slope. Using the chain rule:

d(ln(4*x^2))/d(x)=1/(4*x^2)⋅8*x

d(ln(4*x^2))/d(x)=2/x

7. Interpret the slope for the shape. For x>0 the slope is positive and the function is increasing. For x<0 the slope is negative and the function is decreasing.

Final Answer

The graph of y=ln(4*x^2) is a symmetric curve with a vertical asymptote at x=0 xintercepts at (±0.5,0) and it increases toward infinity as |x| increases.

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