Question 7 Draw neat graphs of the functions on the same set of axes. Clearly indicate intercepts w the axes. $$ f(x)=-x^{2}-2 x+3 $$ $$ g(x)=-2 \cdot 2^{x-1}+1 $$

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Below is one acceptable solution. (Any answer that correctly shows the key features is acceptable.)

We wish to graph on the same axes

f(x) = –x² – 2x + 3    and  g(x) = –2·2^(x–1) + 1

with all intercepts labeled.

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Step 1. Graphing f(x) = –x² – 2x + 3

(a) Rewrite in standard form and find the vertex.  
  f(x) = –(x² + 2x) + 3  
Complete the square inside the parenthesis:  
  x² + 2x = (x + 1)² – 1  
Thus,  
  f(x) = –[(x + 1)² – 1] + 3 = –(x + 1)² + 1 + 3 = –(x + 1)² + 4  
So the vertex is at (–1, 4).

(b) x–intercepts: Solve f(x) = 0  
  –x² – 2x + 3 = 0   Multiply by –1: x² + 2x – 3 = 0  
Factor:  (x + 3)(x – 1) = 0  
Thus, x = –3 and x = 1. So the x–intercepts are (–3,0) and (1,0).

(c) y–intercept: Let x = 0  
  f(0) = –0² – 2·0 + 3 = 3  
So the y–intercept is (0,3).

(d) Plot those points and draw a downward–opening parabola (since the coefficient of (x+1)² is –1).

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Step 2. Graphing g(x) = –2·2^(x–1) + 1

(a) Identify the basic transformation.  
Start with the basic exponential function h(x) = 2^(x). In g(x):

– The factor 2^(x–1) means a horizontal shift 1 unit to the right.
 – The factor –2 means first a vertical stretch by 2 and a reflection in the horizontal (x)–axis.
 – The +1 shifts the graph upward by 1.

(b) Asymptote:  
Because the exponential term 2^(x–1) is always positive, when x → –∞, the term → 0 and  
  g(x) → –2·0 + 1 = 1.  
So there is a horizontal asymptote at y = 1.

(c) Intercepts:  
 • y–intercept: Set x = 0  
  g(0) = –2·2^(0–1) + 1 = –2·2^(–1) + 1 = –2·(1/2) + 1 = –1 + 1 = 0.  
  So the y–intercept is (0,0).

• x–intercept: Set g(x) = 0  
  0 = –2·2^(x–1) + 1  ⟹ 2·2^(x–1) = 1  ⟹ 2^(x–1) = 1/2  
Since 1/2 = 2^(–1), we have:  
  x – 1 = –1 ⟹ x = 0.  
Thus the only intercept is (0,0).

(d) Choose one or two additional points to help plot the curve. For instance, let x = 2:  
  g(2) = –2·2^(2–1) + 1 = –2·2 + 1 = –4 + 1 = –3  
So point (2, –3) lies on the graph.

(e) Plot the asymptote (a dashed line at y = 1), the intercept (0,0) and the point (2, –3). Then sketch the curve with a smooth shape approaching y = 1 from below as x → –∞ and decreasing rapidly as x increases.

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Step 3. Plotting on the Same Axes

1. Draw coordinate axes.
2. For f(x):  
 • Plot the vertex at (–1,4).  
 • Mark the x–intercepts at (–3,0) and (1,0) and the y–intercept at (0,3).  
 • Draw the smooth downward–opening parabola.
3. For g(x):  
 • Draw the horizontal asymptote y = 1 (usually as a dashed line).  
 • Plot the intercept (and only point) (0,0), and another point at (2,–3).  
 • Sketch the exponential curve, keeping in mind that as x → –∞ the graph approaches y = 1 from below, and as x increases the graph falls quickly.

4. Label all the intercepts and the asymptote.

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Final Answer

A correct graph is one that shows:

For f(x) = –x² – 2x + 3:
 • Vertex: (–1, 4)
 • x–intercepts: (–3, 0) and (1, 0)
 • y–intercept: (0, 3)
 • Curve: A downward–opening parabola.

For g(x) = –2·2^(x–1) + 1:
 • Asymptote: y = 1 (dashed line)
 • Intercept: (0, 0) (which is both an x– and y–intercept)
 • Additional point: (2, –3)
 • Curve: An exponential curve that is reflected vertically, shifted right 1 unit, and up 1 unit, so that as x → –∞, g(x) → 1, and then falls quickly for larger x.

Any clear sketch that shows these features – with both curves on the same set of axes and all intercepts labeled – is correct.
To graph the functions $$ f(x) = -x^{2} - 2x + 3 $$ and $$ g(x) = -2 \cdot 2^{x-1} + 1 $$ on the same set of axes, follow these steps: 1. **Graph $$ f(x) = -x^{2} - 2x + 3 $$:** - **Vertex:** At $$ x = -1 $$, $$ f(-1) = 4 $$. So, the vertex is at $$ (-1, 4) $$. - **x-intercepts:** Solve $$ -x^{2} - 2x + 3 = 0 $$ to find $$ x = -3 $$ and $$ x = 1 $$. - **y-intercept:** At $$ x = 0 $$, $$ f(0) = 3 $$. So, the y-intercept is at $$ (0, 3) $$. - **Shape:** Draw a downward-opening parabola passing through these points. 2. **Graph $$ g(x) = -2 \cdot 2^{x-1} + 1 $$:** - **Asymptote:** As $$ x $$ approaches negative infinity, $$ g(x) $$ approaches 1. - **Intercept:** At $$ x = 0 $$, $$ g(0) = 0 $$. So, the only intercept is at $$ (0, 0) $$. - **Additional Point:** At $$ x = 2 $$, $$ g(2) = -3 $$. Plot the point $$ (2, -3) $$. - **Shape:** Sketch an exponential curve that reflects vertically, shifts right by 1 unit, and up by 1 unit, approaching the asymptote $$ y = 1 $$ from below as $$ x $$ increases. 3. **Plotting on the Same Axes:** - Draw the coordinate axes. - Plot all intercepts and key points for both functions. - Draw the asymptote for $$ g(x) $$ as a dashed line. - Sketch both curves ensuring they reflect the described features. **Final Graph:** Both functions should be clearly labeled with their intercepts and the asymptote for $$ g(x) $$ should be indicated.

Question 7 Draw neat graphs of the functions on the same set of axes. Clearly indicate intercepts w the axes. \( f(x)=-x^{2}-2 x+3 \) \( g(x)=-2 \cdot 2^{x-1}+1 \)

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