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Exam (elaborations)

MAT1501 ASSESSMENT 2 SEM 1 OF 2025 (UNIQUE NO 808716)EXPECTED QUESTIONS AND ANSWERS

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THIS DOCUMENT CONTAINS MAT1501 ASSESSMENT 2 SEM 1 OF 2025 EXPECTED QUESTIONS AND ANSWERS. USE IT CORRECTLY AS A GUIDE

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May 14, 2025
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Question 1: Piecewise Function
OSCAR THE TUTOR
1.1 Sketch the graph of g(x)
+27737560989
for FAC MAC ECS DSC TAX QMI FIN INV BNU STA tutoria
The function g(x) is defined as:

⎧ x2 if x < 1
1 if 1 ≤ x < 2
g(x) = ⎨
if 2 ≤ x < 5
​ ​ ​




∣x∣

5 if x ≤ 5

For x < 1: Parabola y = x2 .
For 1 ≤ x < 2: Horizontal line y = 1.
For 2 ≤ x < 5: V-shaped graph y = ∣x∣.
For x ≤ 5: Horizontal line y = 5.

Key Points:

At x = 1: g(1) = 1.
At x = 2: g(2) = 2.
At x = 5: g(5) = 5.

1.2 Why is g a function?

A function assigns exactly one output (y-value) to each input (x-value). For g(x), every x has only one
corresponding y, so it is a function.

1.3 Is g one-to-one?

No, because different x-values can give the same y-value (e.g., g(−1) = 1 and g(1) = 1).

1.4 Equation of f(x) after shifting g(x) upwards by 4 units

⎧ x2 + 4 if x < 1
5 if 1 ≤ x < 2
f(x) = ⎨
∣x∣ + 4 if 2 ≤ x < 5
​ ​ ​





9 if x ≤ 5

, Question 2: Transformations

2.1 Transformation of f(x) = ∣x∣
x x
1. Stretch horizontally by 2: Replace x with 2 : f(x) 2
​ . = ​ ​ ​




2. Shift left by 6 units: Replace x with x + 6: f(x) = x+6
2 .
​ ​ ​




3. Shift downward by 4 units: Subtract 4: f(x) = x+6 2
− 4. ​ ​ ​




Final Equation:

x+6
f(x) = −4
2
​ ​ ​




2.2 Transformation of g(x) = x2
x2
1. Shrink vertically by 16 : Multiply by 16 : g(x)
​ = 6
. ​




2
2. Reflect about the x-axis: Multiply by −1: g(x) = − x6 . ​




2
3. Shift upward by 56 units: Add 56 : g(x)
​ = − x6 + 56 . ​ ​




Final Equation:

x2 5
g(x) = − +
6 6
​ ​




Question 3: Quadratic Function

3.1 Coordinate of B

Given the line 3y = x − 5, and B is the intersection point with f(x). Since B is an x-intercept of f , y = 0:

3(0) = x − 5 ⟹ x = 5

Coordinate of B : (5, 0).

3.2 Equation of f(x)

Given points:

A(−2, 0): 0 = a(−2)2 + b(−2) + c ⟹ 4a − 2b + c = 0.
B(5, 0): 0 = a(5)2 + b(5) + c ⟹ 25a + 5b + c = 0.

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