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11 For which values of \( x \) is \( \delta \) anction: \( g(x)=-\frac{1}{x+2}-1 \) 8.12. 1 Determine the equation of \( v \) if \( v \) is the reflection of \( g \) In the \( y \)-axis. 8.12.2 Determine the symmetry lines of \( v \). 8.12.3 Sketch von a new set of axes, clearly indicating the intercepts with the axes and at least one other point. 8.12.4 Sketch and label the symmetry lines of \( v \). Given that \( w \) is obtained by translating \( g 2 \) units to the right and 1 unit \( u \) 8.13.1 State the equation of \( w \). 8.13.2 State the asymptotes of \( w \). 8.13.3 State the equations of the symmetry lines of \( w \). 8.13.4 Determine the shortest distance between \( w \) and the origin.

Ask by Hall Chandler. in South Africa
Jan 21,2025

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Answer

**Summary of Results:** 1. **Equation of \( v \)**: \( v(x) = -\frac{1}{2-x} - 1 \) 2. **Symmetry Lines of \( v \)**: None 3. **Intercepts of \( v \)**: \( (0, -\frac{3}{2}), (3, 0), (1, -2) \) 4. **Equation of \( w \)**: \( w(x) = -\frac{1}{x} \) 5. **Asymptotes of \( w \)**: \( x = 0 \) (vertical), \( y = 0 \) (horizontal) 6. **Symmetry Lines of \( w \)**: Line \( y = -x \) 7. **Shortest Distance from \( w \) to the Origin**: \( \sqrt{2} \)

Solution

Let's break down the problem step by step. ### 8.12.1 Determine the equation of \( v \) if \( v \) is the reflection of \( g \) in the \( y \)-axis. The function \( g(x) \) is given by: \[ g(x) = -\frac{1}{x+2} - 1 \] To find the reflection of \( g \) in the \( y \)-axis, we replace \( x \) with \( -x \): \[ v(x) = g(-x) = -\frac{1}{-x+2} - 1 \] Now, simplifying \( v(x) \): \[ v(x) = -\frac{1}{2-x} - 1 \] ### 8.12.2 Determine the symmetry lines of \( v \). To find the symmetry lines of \( v \), we check if \( v(x) \) is even or odd. A function is even if \( v(x) = v(-x) \) and odd if \( v(-x) = -v(x) \). Calculating \( v(-x) \): \[ v(-x) = -\frac{1}{2+x} - 1 \] Now, we check if \( v(-x) = -v(x) \): \[ -v(x) = \frac{1}{2-x} + 1 \] Since \( v(-x) \neq -v(x) \) and \( v(x) \neq v(-x) \), the function \( v \) is neither even nor odd. Therefore, there are no symmetry lines. ### 8.12.3 Sketch \( v \) on a new set of axes, clearly indicating the intercepts with the axes and at least one other point. To find the intercepts: 1. **Y-intercept**: Set \( x = 0 \): \[ v(0) = -\frac{1}{2} - 1 = -\frac{3}{2} \] So, the y-intercept is \( (0, -\frac{3}{2}) \). 2. **X-intercept**: Set \( v(x) = 0 \): \[ -\frac{1}{2-x} - 1 = 0 \] Solving for \( x \): \[ -\frac{1}{2-x} = 1 \implies -1 = 2 - x \implies x = 3 \] So, the x-intercept is \( (3, 0) \). 3. **Another point**: Let's calculate \( v(1) \): \[ v(1) = -\frac{1}{2-1} - 1 = -1 - 1 = -2 \] So, another point is \( (1, -2) \). ### 8.12.4 Sketch and label the symmetry lines of \( v \). Since there are no symmetry lines, we will only sketch the function \( v \) with the intercepts and the additional point. ### 8.13.1 State the equation of \( w \). The function \( w \) is obtained by translating \( g \) 2 units to the right and 1 unit up. The transformation can be expressed as: \[ w(x) = g(x-2) + 1 \] Calculating \( g(x-2) \): \[ g(x-2) = -\frac{1}{(x-2)+2} - 1 = -\frac{1}{x} - 1 \] Thus, \[ w(x) = -\frac{1}{x} - 1 + 1 = -\frac{1}{x} \] ### 8.13.2 State the asymptotes of \( w \). The function \( w(x) = -\frac{1}{x} \) has: - A vertical asymptote at \( x = 0 \). - A horizontal asymptote at \( y = 0 \). ### 8.13.3 State the equations of the symmetry lines of \( w \). The function \( w(x) = -\frac{1}{x} \) is an odd function, so it has symmetry about the origin. Therefore, the symmetry line is: - The line \( y = -x \). ### 8.13.4 Determine the shortest distance between \( w \) and the origin. The shortest distance from the origin to the curve \( w(x) = -\frac{1}{x} \) occurs when the distance function is minimized. The distance \( d \) from the origin to a point \( (x, w(x)) \) is given by: \[ d = \sqrt{x^2 + w(x)^2} = \sqrt{x^2 + \left(-\frac{1}{x}\right)^2} = \sqrt{x^2 + \frac{1}{x^2}} \] To minimize \( d \), we can minimize \( d^2 \): \[ d^2 = x^2 + \frac{1}{x^2} \] Taking the derivative and setting it to zero: \[ \frac{d}{dx}(x^2 + \frac{1}{x^2}) = 2x - \frac{2}{x^3} = 0 \] Solving for \( x \): \[ 2x^4 = 2 \implies x^4 = 1 \implies x = 1 \text{ (since distance cannot be negative)} \] Now substituting \( x = 1 \) back into \( w(x) \): \[ w(1) = -1 \] Thus, the point is \( (1, -1) \) and the distance is: \[ d = \sqrt{1^2 + (-1)^2} = \sqrt{2} \] ### Summary of Results: - **Equation of \( v \)**: \( v(x) = -\frac{1}{2-x} - 1 \) - **Symmetry lines of \( v \)**: None - **Intercepts of \( v \)**: \( (0, -\frac{3}{2}), (3, 0), (1, -2) \) - **Equation of \( w \)**: \( w(x) = -\frac{1}{x} \) - **Asymptotes of \( w \)**: \( x = 0 \) (vertical), \( y = 0 \) (horizontal) - **Symmetry lines of \( w \)**: Line \( y = -x \) - **Shortest distance from \( w \) to the origin**: \( \sqrt{2} \)

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The Deep Dive

To reflect the function \( g(x) = -\frac{1}{x+2} - 1 \) across the \( y \)-axis, you’ll want to replace \( x \) with \( -x \). So, the equation becomes \( v(x) = -\frac{1}{-x+2} - 1 = -\frac{1}{2-x} - 1 \). This transformed equation represents your reflection, so get your sketchbook ready for some graphing fun! For \( w(x) \), since it involves translating \( g \) 2 units to the right and 1 unit up, we adjust the original function accordingly. Thus, the equation becomes \( w(x) = -\frac{1}{(x-2)+2} - 1 + 1 = -\frac{1}{x} \). As for the asymptotes of \( w \), we can see that the vertical asymptote occurs at \( x = 0 \) since the function is undefined there, and the horizontal asymptote remains at \( y = 0 \). Now you’re all set to analyze transformations and asymptotes like a pro!

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