\(y = -3(x - 5)^2 + 1\)
Value of \(a\):
\(a\) is: ( Positive / Negative )
Opens: ( Upward / Downward )
\(y = \frac{2}{5}x^2 - 8x + 3\)
Value of \(a\):
\(a\) is: ( Positive / Negative )
Opens: ( Upward / Downward )
\(y = -\frac{7}{3}(x + 2)(x - 6)\)
Value of \(a\):
\(a\) is: ( Positive / Negative )
Opens: ( Upward / Downward )
\(y = 4(x + 1)(x + 7)\)
Value of \(a\):
\(a\) is: ( Positive / Negative )
Opens: ( Upward / Downward )
\(y = -0.75x^2 + 12\)
Value of \(a\):
\(a\) is: ( Positive / Negative )
Opens: ( Upward / Downward )
\(y = \frac{1}{6}(x - 4)^2 - \frac{8}{3}\)
Value of \(a\):
\(a\) is: ( Positive / Negative )
Opens: ( Upward / Downward )
• Determine which feature is missing between Directrix, Focus, and Vertex.
• Write the coordinates of the vertex and focus
• Sketch the missing feature.
• Write the equation of the axis of symmetry and the directrix
• Sketch the parabola opening up or down.
• Write the value of p
17. Given: Vertex & Focus
x y Vertex Focus
Vertex:
Focus:
Axis of Sym:
Directrix:
p-value:
18. Given: Vertex & Directrix
x y Directrix Vertex
Vertex:
Focus:
Axis of Sym:
Directrix:
p-value:
19. Given: Focus & Directrix
x y Focus Directrix
Vertex:
Focus:
Axis of Sym:
Directrix:
p-value:
20. Given: Vertex & Focus
x y Vertex Focus
Vertex:
Focus:
Axis of Sym:
Directrix:
p-value:
A. Technology Graphing (Pure Mathematical Equations)
Graph each quadratic equation using your graphing calculator or graphing software. State the vertex coordinate, opening direction, domain, and range.
\(y = 3(x - 2)^2 - 8\)
Vertex coordinate:
Opens: ( Upward / Downward )
Domain:
Range:
\(y = -\frac{1}{2}(x + 4)^2 + 10\)
Vertex coordinate:
Opens: ( Upward / Downward )
Domain:
Range:
B. Contextual Word Problems (Physical Reality Constraints)
Real-world situations place natural physical boundaries on variables (such as time \(\ge 0\) and height above the ground). Determine the realistic domain and range.
23. High Dive Scenario
A cliff diver jumps from a platform into the ocean. Her height \(h(t)\) in feet above the water is given by \(h(t) = -16t^2 + 8t + 24\), where \(t\) represents elapsed time in seconds. She reaches a maximum height of 25 feet at \(t = 0.25\) seconds and enters the water (\(h = 0\)) at \(t = 1.5\) seconds.
Realistic Domain:
Realistic Range:
Explanation: Why is the domain of this scenario restricted instead of being all real numbers?
24. Model Rocket Flight
A science class launches a model rocket from ground level (\(h = 0\)). Its altitude \(h(t)\) in meters after \(t\) seconds is modeled by the function \(h(t) = -5t^2 + 40t\). The rocket reaches its peak altitude of 80 meters at \(t = 4\) seconds and parachutes back to the ground at \(t = 8\) seconds.
Realistic Domain:
Realistic Range:
Explanation: Explain what physical event sets the upper bound of the domain:
Solved Example
Position: Inside / Outside Type: Horizontal / Vertical Vertical reflection Position: Inside / Outside Type: Horizontal / Vertical Shift right 3 units y = − 2 (x − 3 )² + 5 Position: Inside / Outside Type: Horizontal / Vertical Vertical stretch by a factor of 2 Position: Inside / Outside Type: Horizontal / Vertical Shift up 5 units
Problem 25
Position: ( Inside / Outside ) Type: ( Horizontal / Vertical ) DESCRIBE TRANSFORMATION: Position: ( Inside / Outside ) Type: ( Horizontal / Vertical ) DESCRIBE TRANSFORMATION: y = 2 (x − 4 )² + 3 Position: ( Inside / Outside ) Type: ( Horizontal / Vertical ) DESCRIBE TRANSFORMATION:
Problem 26
Position: ( Inside / Outside ) Type: ( Horizontal / Vertical ) DESCRIBE TRANSFORMATION: Position: ( Inside / Outside ) Type: ( Horizontal / Vertical ) DESCRIBE TRANSFORMATION: y = − 1 2 (x + 5 )² − 6 Position: ( Inside / Outside ) Type: ( Horizontal / Vertical ) DESCRIBE TRANSFORMATION: Position: ( Inside / Outside ) Type: ( Horizontal / Vertical ) DESCRIBE TRANSFORMATION:
Problem 27
y = 3(x + 2)² − 8
Problem 28
y = −
1 4
(x − 6)² + 7
