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Sketching and Finding the Exercise 5: Sketch the graph a) \( y=2^{x+1}+2 \) b) \( f(x)=-(2)^{x+1}-2 \)

Ask by Vaughan Salinas. in South Africa
Feb 02,2025

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To sketch the graphs: **a) \( y = 2^{x+1} + 2 \):** - Shift the basic \( y = 2^x \) graph 1 unit left and 2 units up. - As \( x \) increases, the graph rises exponentially. - As \( x \) decreases, it approaches the horizontal asymptote \( y = 2 \). **b) \( f(x) = -2^{x+1} - 2 \):** - Reflect the basic \( y = 2^x \) graph over the x-axis and shift it 1 unit left and 2 units down. - As \( x \) increases, the graph decreases exponentially. - As \( x \) decreases, it approaches the horizontal asymptote \( y = -2 \). Plot key points and draw the curves accordingly.

Solution

Sure, let's analyze and describe how to sketch the graphs for both functions: ### Exercise 5: Sketch the Graph #### a) \( y = 2^{x+1} + 2 \) **1. Basic Function:** - Start with the basic exponential function \( y = 2^x \). **2. Horizontal Shift:** - The term \( x + 1 \) inside the exponent indicates a **horizontal shift**. - Specifically, \( y = 2^{x+1} \) shifts the graph **1 unit to the left**. **3. Vertical Shift:** - Adding 2 shifts the entire graph **2 units upward**. - So, \( y = 2^{x+1} + 2 \) moves the graph of \( y = 2^x \) left by 1 and up by 2. **4. Asymptote:** - The horizontal asymptote for \( y = 2^{x+1} + 2 \) is \( y = 2 \). **5. Key Points:** - Choose some x-values to plot: - When \( x = -1 \): \[ y = 2^{-1+1} + 2 = 2^0 + 2 = 1 + 2 = 3 \] Point: (-1, 3) - When \( x = 0 \): \[ y = 2^{0+1} + 2 = 2^1 + 2 = 2 + 2 = 4 \] Point: (0, 4) - When \( x = 1 \): \[ y = 2^{1+1} + 2 = 2^2 + 2 = 4 + 2 = 6 \] Point: (1, 6) **6. Sketch Summary:** - **Start**: The graph approaches the asymptote \( y = 2 \) as \( x \) approaches negative infinity. - **Increasing**: As \( x \) increases, the graph rises exponentially. - **Key Points**: Plot the points calculated above and draw a smooth curve passing through them, getting closer to \( y = 2 \) on the left and rising steeply to the right. --- #### b) \( f(x) = -2^{x+1} - 2 \) **1. Basic Function:** - Start with the basic exponential function \( y = 2^x \). **2. Horizontal Shift:** - Similar to part (a), \( x + 1 \) inside the exponent indicates a **horizontal shift** of **1 unit to the left**. - So, \( y = 2^{x+1} \). **3. Reflection over the x-axis:** - The negative sign in front of \( 2^{x+1} \) reflects the graph over the **x-axis**. - Thus, \( y = -2^{x+1} \) flips the graph vertically. **4. Vertical Shift:** - Subtracting 2 shifts the entire graph **2 units downward**. - Therefore, \( f(x) = -2^{x+1} - 2 \) moves the reflected graph left by 1 and down by 2. **5. Asymptote:** - The horizontal asymptote for \( f(x) = -2^{x+1} - 2 \) is \( y = -2 \). **6. Key Points:** - Choose some x-values to plot: - When \( x = -1 \): \[ f(-1) = -2^{-1+1} - 2 = -2^0 - 2 = -1 - 2 = -3 \] Point: (-1, -3) - When \( x = 0 \): \[ f(0) = -2^{0+1} - 2 = -2^1 - 2 = -2 - 2 = -4 \] Point: (0, -4) - When \( x = 1 \): \[ f(1) = -2^{1+1} - 2 = -2^2 - 2 = -4 - 2 = -6 \] Point: (1, -6) **7. Sketch Summary:** - **Start**: The graph approaches the asymptote \( y = -2 \) as \( x \) approaches negative infinity. - **Decreasing**: As \( x \) increases, the graph decreases exponentially. - **Key Points**: Plot the points calculated above and draw a smooth curve passing through them, getting closer to \( y = -2 \) on the left and descending steeply to the right. --- ### Visual Comparison: - **Function (a)** is an exponential growth curve shifted left by 1 and up by 2, with an asymptote at \( y = 2 \). - **Function (b)** is a reflected exponential decay curve (due to the negative sign) shifted left by 1 and down by 2, with an asymptote at \( y = -2 \). By plotting the key points and considering the shifts and reflections, you can accurately sketch both graphs.

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To sketch the graph of \( y=2^{x+1}+2 \), start by recognizing that the basic shape of \( 2^x \) is an upward-opening curve. The \( +1 \) in the exponent shifts the graph to the left by 1 unit, and the \( +2 \) moves it up 2 units. At \( x = 0 \), \( y = 2^{1} + 2 = 4 \). The horizontal asymptote is at \( y = 2 \), and as \( x \) heads towards negative infinity, the graph approaches this line. For the graph of \( f(x)=-(2)^{x+1}-2 \), we can see that it is a reflection of the previous function across the x-axis due to the negative sign. The \( +1 \) again shifts the base curve left by 1 unit, while the \( -2 \) shifts it down by 2 units. At \( x = 0 \), \( f(0) = -2^{1} - 2 = -4 \). The horizontal asymptote here is at \( y = -2 \), and as \( x \) approaches negative infinity, the curve approaches this line.

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