Question

use a graphing calculator to solve the equation. round your answer to two decimal places. ( e^x = -x ) ( \bigcirc ) -1.05 ( \bigcirc ) 1.05 ( \bigcirc ) -0.57 ( \bigcirc ) 0.57 clear my selection
18 multiple choice 4 points graph the function. ( f(x) = \frac{2x}{(x - 2)(x + 3)} ) (with y - axis up to 40)

Explanation:

Problem 1 (Solving \( e^x = -x \))

Step1: Analyze the equation

We need to find the \( x \)-value where the functions \( y = e^x \) and \( y=-x \) intersect. Using a graphing calculator, we can graph both functions and find their intersection point.

Step2: Use graphing calculator

When we graph \( y = e^x \) (an exponential growth function) and \( y=-x \) (a linear function with slope -1 and y-intercept 0), we observe their intersection. By using the intersection feature of the graphing calculator, we find that the \( x \)-coordinate of the intersection point is approximately -0.57 (after rounding to two decimal places).

Step1: Identify key features

• Vertical Asymptotes: Set the denominator equal to zero: \( (x - 2)(x + 3)=0 \). So, vertical asymptotes at \( x = 2 \) and \( x=-3 \).
• Horizontal Asymptote: The degree of the numerator (1) is less than the degree of the denominator (2), so the horizontal asymptote is \( y = 0 \) (the x-axis).
• x-intercept: Set the numerator equal to zero: \( 2x=0\implies x = 0 \). So, the x-intercept is at \( (0,0) \), which is also the y-intercept (since when \( x = 0 \), \( f(0)=0 \)).
• Sign Analysis:
• For \( x < -3 \): Let's take \( x=-4 \). \( f(-4)=\frac{2(-4)}{(-4 - 2)(-4 + 3)}=\frac{-8}{(-6)(-1)}=\frac{-8}{6}<0 \).
• For \( -3 < x < 0 \): Let's take \( x=-1 \). \( f(-1)=\frac{2(-1)}{(-1 - 2)(-1 + 3)}=\frac{-2}{(-3)(2)}=\frac{-2}{-6}>0 \).
• For \( 0 < x < 2 \): Let's take \( x = 1 \). \( f(1)=\frac{2(1)}{(1 - 2)(1 + 3)}=\frac{2}{(-1)(4)}=\frac{2}{-4}<0 \).
• For \( x > 2 \): Let's take \( x = 3 \). \( f(3)=\frac{2(3)}{(3 - 2)(3 + 3)}=\frac{6}{(1)(6)} = 1>0 \).

Step2: Sketch the graph

• Draw the vertical asymptotes \( x=-3 \) and \( x = 2 \) as dashed lines.
• Draw the horizontal asymptote \( y = 0 \) (the x-axis).
• Plot the x-intercept (and y-intercept) at \( (0,0) \).
• Use the sign analysis to sketch the branches of the graph:
• For \( x < -3 \), the graph is below the x-axis, approaching the vertical asymptote \( x=-3 \) from the left and the horizontal asymptote \( y = 0 \) as \( x\to-\infty \).
• For \( -3 < x < 0 \), the graph is above the x-axis, passing through \( (0,0) \) and approaching the vertical asymptote \( x=-3 \) from the right.
• For \( 0 < x < 2 \), the graph is below the x-axis, approaching the vertical asymptote \( x = 2 \) from the left.
• For \( x > 2 \), the graph is above the x-axis, approaching the vertical asymptote \( x = 2 \) from the right and the horizontal asymptote \( y = 0 \) as \( x\to\infty \).

(Note: Since the problem asks to graph the function, the above steps describe how to construct the graph. If a visual graph is needed, it can be plotted using the key features identified.)

Answer:

[-0.57]

Problem 2 (Graphing \( f(x)=\frac{2x}{(x - 2)(x + 3)} \))