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QUESTION 3
The equation of a hyperbola is given by .
Write down the equation of the new function that is forme
Shift two units to the left
Shift 3 units up
Shift 1 unit right and 2 units down
The equation of the new hyperbola has new asyr
QUESTION 4
Sketch on the same set of axes the graphs of .
Clearly indicate all intercepts with the axes, turning point

Ask by Burton Vargas. in South Africa
Jan 08,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

QUESTION 3
  1. Shift Two Units to the Left:
  2. Shift Three Units Up:
  3. Shift One Unit Right and Two Units Down:
  4. Asymptotes After Transformations:
    • After 3.1: Vertical asymptote at , horizontal asymptote at
    • After 3.2: Vertical asymptote at , horizontal asymptote at
    • After 3.3: Vertical asymptote at , horizontal asymptote at
QUESTION 4
  • Graph of :
    • Vertex: (0, 0)
    • Y-intercept: (0, 0)
    • X-intercept: (0, 0)
    • Symmetric about the y-axis
    • Opens downward
    • Additional Points: (1, -2) and (-1, -2)
By plotting these points and understanding the symmetry, the graph of can be accurately sketched.

Solution

QUESTION 3

Given the hyperbola , we’ll apply various transformations as specified.

3.1 Shift Two Units to the Left

Transformation Steps:
  • Horizontal Shift: Shifting the graph two units to the left involves replacing with .
New Function:
Final Equation:

3.2 Shift Three Units Up

Transformation Steps:
  • Vertical Shift: Shifting the graph three units up involves adding 3 to the function.
New Function:
Final Equation:

3.3 Shift One Unit Right and Two Units Down

Transformation Steps:
  • Horizontal Shift: Shifting the graph one unit to the right involves replacing with .
  • Vertical Shift: Shifting the graph two units down involves subtracting 2 from the function.
New Function:
Final Equation:

3.4 New Asymptotes of the Transformed Hyperbola

For the transformed hyperbolas obtained in parts 3.1 to 3.3, the asymptotes change based on the transformations applied.
  1. Original Function :
    • Vertical Asymptote:
    • Horizontal Asymptote:
  2. After 3.1 :
    • Vertical Asymptote:
    • Horizontal Asymptote:
  3. After 3.2 :
    • Vertical Asymptote:
    • Horizontal Asymptote:
  4. After 3.3 :
    • Vertical Asymptote:
    • Horizontal Asymptote:

QUESTION 4

Sketching the Graph of :
Since a sketch cannot be drawn here, I’ll describe the key features of the graph:
  1. Shape of the Graph:
    • The equation represents a downward-opening parabola because the coefficient of is negative.
  2. Vertex (Turning Point):
    • The vertex of the parabola is at the origin.
  3. Intercepts:
    • Y-intercept: Set .
      The y-intercept is at the origin.
    • X-intercepts: Set .
      The only x-intercept is also at the origin.
  4. Additional Points:
    • For :
    • For :
    • These points help in sketching the symmetry of the parabola about the y-axis.
  5. Symmetry:
    • The parabola is symmetric with respect to the y-axis.
  6. Behavior:
    • As increases, decreases without bound due to the negative coefficient.
Summary of Key Features:
  • Vertex: (0, 0)
  • Y-intercept: (0, 0)
  • X-intercept: (0, 0)
  • Opening Direction: Downward
  • Symmetry: About the y-axis
By plotting these points and understanding the symmetry, one can sketch an accurate graph of the parabola .

Answered by UpStudy AI and reviewed by a Professional Tutor

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Simplify this solution

Beyond the Answer

To shift the hyperbola two units to the left, you need to replace with . So, the new function becomes:
.
For a 3-unit upward shift, simply add 3 to the entire function, resulting in:
.
By shifting 1 unit to the right and 2 units down, replace with and subtract 2:
Thus, the equation becomes:
.
The new asymptotes will change as you perform these transformations. The original had vertical asymptote at . After the shifts, the vertical asymptote for each transformation adjusts accordingly.

The graph of is a downward-opening parabola with its vertex at the origin (0,0). The x-intercepts occur where , giving as the only intercept, while the y-intercept is also at (0,0). The parabola has no turning points other than the vertex since it continuously decreases. The sketch should illustrate the origin and show the arms of the parabola extending downwards.

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