Question: User: Quarter 1 Exam
Mathematics 9
Name :____________________________________ Score: _______________ Grade Level: _______________________________ Date: ________________
I. Multiple Choice
Direction: Choose the letter of the best answer and write it on the space provided before each
number. 1. It is a polynomial equation of degree two that can be written in the
format: x^2 + bx + c = 0, where a, b, and c are real numbers and a \neq 0. a. Linear Equation c. Quadratic Equation
b. Linear Inequality d. Quadratic Inequality
2. Which of the following is a quadratic equation?
a. 3s^2 + s - 4 c. 2x - 1 = 5
b. m^2 - 8m - 1 = 0 d. 5y^2 7
3. In the quadratic equation 2x^2 - 9x - 5 = 0, which is the quadratic
term?
a. 2x^2 c. -9x
b. x^2 d. -5
4. In the quadratic equation 2x^2 - 9x - 5 = 0, which is the linear
term?
a. 2x^2 c. -9x
b. x^2 d. -5
5. In the equation 2x^2 - 9x - 5 = 0, which is the constant term?
a. -9x c. 2x^2
b. 5 d. -5
6. _____ is the turning point of the parabola. a. Axis of symmetry c. Intercept
b. Range d. Vertex
7. It is the imaginary line that passes through the vertex. a. Axis of symmetry c. Direction of Opening
b. Domain d. Range
8. What is the axis of symmetry of the function f(x)=-2x^2 -8x -3?
a. -2 c. -3
b. 5 d. 8
9. Determine the opening of the function y=64-x^2.
a. Downward c. Left
b. Right d. Upward
10. Determine the opening of the function y=64+x^2.
a. Downward c. Left
b. Right d. Upward
II. True or False
Direction: Write TRUE if the statement is true and FALSE if it is not. Write your answer on the space
provided before each number. __________ 1. First degree equation is called Linear Equation. __________ 2. Second degree equation is called Quadratic Equation. _________ 3. x + y is a Linear Equation^2
_________ 4. The equation x^2+2x=0 has only one solution. 5. A quadratic equation can have at most two real roots. 6. The graph of a quadratic equation is a parabola. 7. All quadratic equations can be solved by factoring. 8. The standard form of a quadratic equation is ax^2 +bx+c=0, where a=0. 9. If the leading coefficient a is positive, the parabola opens downwards. 10. The quadratic formula is used to find the solutions to a quadratic equation. III. Solving
Direction: Perform the needed operation(s) and simplify your answer. Show your complete solution. (50 points)
1. 4x^2 + 8x + 5 = 1 (Solve for x by using at least any 2 methods, then check)
2. Graph the function y=x^2+2x+1 and label its vertex, axis of symmetry, x-intercepts, and y-intercept. Prepared by: EMANUEL V. WAING
Teacher, Math 9
3
This Table of Specification (TOS) organizes your exam on Quadratic Equations and
Functions. The items are categorized based on Bloom’s Revised Taxonomy, moving from
identifying basic terms to the synthesis required for graphing and multi-method solving. Table of Specification (TOS)
Topic / Learning Competency
Remembering
Understanding
Applying
Analyzing
Evaluating
Creating
Total Items
Definitions & Identification (Quadratic vs. Linear, Terms)
I-1, I-3, I-4, I-5
I-2; II-1, II-2, II-3, II-8 9
Properties of Parabolas (Vertex, Axis, Opening, Range)
I-6, I-7 II-6, II-9
I-8, I-9, I-10 7
Nature of Roots & Solving (Quadratic Formula, Solutions)
II-10 II-5 II-4, II-7
IV-1 (Check)
IV-1 (Solve) 6
Graphing Quadratic Functions (Intercepts, Vertex, Sketching)
IV-2 1
TOTAL ITEMS 6 9 3 2 1 2
1. **Problem:** Solve the quadratic equation $$4x^2 + 8x + 5 = 1$$ for $$x$$ using at least two methods and check the solutions.
2. **Step 1: Write the equation in standard form.**
$$4x^2 + 8x + 5 = 1 \implies 4x^2 + 8x + 5 - 1 = 0$$
$$4x^2 + 8x + 4 = 0$$
3. **Step 2: Simplify the equation by dividing all terms by the greatest common factor, which is 4.**
$$\cancel{4}x^2 + \cancel{4}2x + \cancel{4} = 0 \implies x^2 + 2x + 1 = 0$$
4. **Method 1: Factoring**
Recognize that $$x^2 + 2x + 1$$ is a perfect square trinomial.
$$x^2 + 2x + 1 = (x + 1)^2 = 0$$
Set the factor equal to zero:
$$(x + 1)^2 = 0 \implies x + 1 = 0 \implies x = -1$$
5. **Method 2: Quadratic Formula**
Recall the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
For the equation $$x^2 + 2x + 1 = 0$$, the coefficients are $$a=1$$, $$b=2$$, and $$c=1$$.
Calculate the discriminant:
$$\Delta = b^2 - 4ac = 2^2 - 4(1)(1) = 4 - 4 = 0$$
Since $$\Delta = 0$$, there is one real root.
Calculate $$x$$:
$$x = \frac{-2 \pm \sqrt{0}}{2(1)} = \frac{-2}{2} = -1$$
6. **Step 3: Check the solution by substituting $$x = -1$$ back into the original equation:**
$$4(-1)^2 + 8(-1) + 5 = 4(1) - 8 + 5 = 4 - 8 + 5 = 1$$
The left side equals the right side, so $$x = -1$$ is a valid solution.
**Final answer:** $$x = -1$$ (one real root, repeated)