Gina Wilson All Things Algebra 2014 Unit 4 Homework 3

Gina Wilson – All Things Algebra (2014) – Unit 4 Homework 3: A Complete Guide

Introduction

All Things Algebra by Gina Wilson has been a staple for high‑school students since its first edition, and the 2014 revision remains one of the most widely used textbooks for Algebra I and Algebra II courses. Unit 4 Homework 3 is the third assignment in the unit that focuses on quadratic functions, factoring techniques, and the application of the quadratic formula. This article breaks down every problem type found in the worksheet, explains the underlying concepts, provides step‑by‑step solutions, and offers tips for mastering the skills required to score full credit. By the end of this guide, you will not only have the answers you need but also a deeper conceptual understanding that will serve you throughout the rest of the course.

1. What Unit 4 Covers

Before diving into the homework, it helps to review the core objectives of Unit 4:

1. Identify the standard form of a quadratic function – (f(x)=ax^{2}+bx+c).
2. Graph quadratic functions using vertex, axis of symmetry, and intercepts.
3. Factor quadratics by grouping, using the “ac‑method,” and recognizing perfect square trinomials.
4. Solve quadratic equations by factoring, completing the square, and applying the quadratic formula.
5. Interpret real‑world problems that model situations with quadratic relationships (projectile motion, area optimization, etc.).

Homework 3 is designed to assess each of these objectives, with a mix of computational and word‑problem questions.

2. Problem‑by‑Problem Walkthrough

Below is a detailed analysis of the typical questions found on the worksheet. The numbering follows the original PDF; if your copy differs slightly, the strategies still apply.

Problem 1 – Converting to Standard Form

Write the quadratic function whose vertex is ((3,,-2)) and passes through the point ((5,6)).

Solution Steps

1. Start with vertex form: (f(x)=a(x-h)^{2}+k) where ((h,k)=(3,-2)).
   [ f(x)=a(x-3)^{2}-2 ]
2. Plug the given point ((5,6)) to find (a):
   [ 6=a(5-3)^{2}-2;\Rightarrow;6=a(2)^{2}-2;\Rightarrow;6=4a-2 ]
   [ 4a=8;\Rightarrow;a=2 ]
3. Write the function in vertex form: (f(x)=2(x-3)^{2}-2).
4. Expand to standard form:
   [ f(x)=2(x^{2}-6x+9)-2=2x^{2}-12x+18-2=2x^{2}-12x+16 ]

Answer: (f(x)=2x^{2}-12x+16).

Tip: Always verify the point after solving for (a); a simple arithmetic slip can derail the entire problem.

Problem 2 – Factoring by the “ac‑Method”

Factor completely: (6x^{2}+11x-35).

Solution Steps

1. Multiply (a) and (c): (6 \times (-35) = -210).
2. Find two numbers that multiply to (-210) and add to (b=11). The pair is 21 and ‑10 because (21 \times (-10) = -210) and (21 + (-10) = 11).
3. Rewrite the middle term:
   [ 6x^{2}+21x-10x-35 ]
4. Factor by grouping:
   [ 3x(2x+7)-5(2x+7) ]
5. Pull out the common binomial:
   [ (2x+7)(3x-5) ]

Answer: ((2x+7)(3x-5)).

Tip: When the “ac‑method” feels cumbersome, list factor pairs of (|ac|) on a small table; it speeds up the search for the correct pair.

Problem 3 – Solving Quadratics with the Quadratic Formula

Solve (4x^{2}-12x+9=0) and state whether the solutions are rational, irrational, or complex.

Solution Steps

1. Identify coefficients: (a=4), (b=-12), (c=9).
2. Compute the discriminant: (\Delta=b^{2}-4ac=(-12)^{2}-4(4)(9)=144-144=0).
3. Apply the quadratic formula:
   [ x=\frac{-b\pm\sqrt{\Delta}}{2a}= \frac{12\pm 0}{8}= \frac{12}{8}= \frac{3}{2} ]
4. Because (\Delta=0), there is one repeated rational root.

Answer: (x=\dfrac{3}{2}) (a rational, double root).

Tip: A zero discriminant always signals a perfect square trinomial; you could factor directly as ((2x-3)^{2}=0).

Problem 4 – Completing the Square

Rewrite (x^{2}+6x-7) in vertex form.

Solution Steps

1. Isolate the constant term: (x^{2}+6x = 7).
2. Take half of the coefficient of (x) (which is (6)), square it: ((6/2)^{2}=9).
3. Add and subtract 9 inside the expression:
   [ x^{2}+6x+9-9-7 = (x+3)^{2}-16 ]
4. Hence, vertex form is ((x+3)^{2}-16).

Answer: (f(x)=(x+3)^{2}-16).

Tip: Keep the added square term balanced by subtracting it immediately; this prevents errors when moving constants to the other side.

Problem 5 – Graph Interpretation

The graph of (y=-2x^{2}+8x-6) intersects the x‑axis at two points. Find the x‑coordinates of these intercepts without using a calculator.

Solution Steps

1. Set (y=0): (-2x^{2}+8x-6=0).
2. Divide by (-2) to simplify: (x^{2}-4x+3=0).
3. Factor: ((x-1)(x-3)=0).
4. Solve for (x): (x=1) or (x=3).

Answer: The x‑intercepts are at (x=1) and (x=3).

Tip: Always check whether the leading coefficient is negative; dividing by it can turn a cumbersome quadratic into a much easier one.

Problem 6 – Real‑World Application

A ball is thrown upward from a 2‑meter platform with an initial velocity of 12 m/s. Its height after (t) seconds is given by (h(t) = -5t^{2}+12t+2).
a) When does the ball reach its maximum height?
b) What is that maximum height?
c) When does the ball hit the ground?

Solution Steps

a) Time of maximum height – occurs at the vertex of the parabola. For (h(t)=at^{2}+bt+c) with (a=-5) and (b=12):
[ t_{\text{vertex}} = -\frac{b}{2a}= -\frac{12}{2(-5)} = \frac{12}{10}=1.2\text{ s} ]

b) Maximum height – plug (t=1.2) back into the equation:
[ h(1.2)= -5(1.2)^{2}+12(1.2)+2 = -5(1.44)+14.4+2 = -7.2+16.4 = 9.2\text{ m} ]

c) When the ball hits the ground – solve (h(t)=0):
[ -5t^{2}+12t+2=0 \quad\Rightarrow\quad 5t^{2}-12t-2=0 ]
Use the quadratic formula:
[ t=\frac{12\pm\sqrt{(-12)^{2}-4(5)(-2)}}{2(5)}=\frac{12\pm\sqrt{144+40}}{10}=\frac{12\pm\sqrt{184}}{10} ]
[ \sqrt{184}\approx13.56 ]
Positive root:
[ t=\frac{12+13.56}{10}= \frac{25.56}{10}=2.556\text{ s} ]
(The negative root is extraneous for time.)

Answers:
a) (t=1.2) s, b) maximum height (9.2) m, c) ball lands at (t\approx2.56) s.

Tip: When the quadratic coefficient is negative (opening downward), the vertex gives the maximum; if it were positive, the vertex would give the minimum.

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Problem 7 – Transformations of Quadratic Graphs

Describe the transformation that changes (y=x^{2}) into (y= -3(x-4)^{2}+5).

Explanation

1. Horizontal shift: ((x-4)) moves the graph right 4 units.
2. Vertical stretch/compression & reflection: The factor (-3) does three things:
   • The negative sign reflects the parabola across the x‑axis (opens downward).
   • The magnitude (3) stretches the graph vertically by a factor of 3 (makes it narrower).
3. Vertical shift: The “+5” lifts the entire graph up 5 units.

Summary: Right 4, reflect over the x‑axis, vertical stretch by 3, then up 5.

Tip: Perform transformations in the order horizontal shift → vertical stretch/reflection → vertical shift to avoid confusion.

Problem 8 – Determining the Nature of Roots

For which values of (k) does the equation (x^{2}+kx+9=0) have two distinct real roots?

Solution Steps

1. Real, distinct roots require a positive discriminant: (\Delta = k^{2} - 4(1)(9) > 0).
2. Simplify: (k^{2} - 36 > 0).
3. Solve the inequality:
   [ k^{2} > 36 ;\Rightarrow; |k| > 6 ]
   Hence, (k < -6) or (k > 6).

Answer: (k\in (-\infty,-6)\cup(6,\infty)).

Tip: Remember that the discriminant tells you everything about root nature:

• (\Delta>0) → two distinct real roots,
• (\Delta=0) → one repeated real root,
• (\Delta<0) → two complex conjugates.

Problem 9 – Polynomial Long Division (Optional Challenge)

Divide (2x^{3}+3x^{2}-5x+6) by (x-2).

Solution Steps

1. Set up the long division.
2. (2x^{3}\div x = 2x^{2}). Multiply back: (2x^{2}(x-2)=2x^{3}-4x^{2}). Subtract: ((3x^{2}+4x^{2})=7x^{2}).
3. Bring down (-5x): (7x^{2}\div x = 7x). Multiply: (7x(x-2)=7x^{2}-14x). Subtract: ((-5x+14x)=9x).
4. Bring down (+6): (9x\div x = 9). Multiply: (9(x-2)=9x-18). Subtract: (6-(-18)=24).

Quotient: (2x^{2}+7x+9) with remainder (24).

Answer: (\displaystyle \frac{2x^{3}+3x^{2}-5x+6}{x-2}=2x^{2}+7x+9+\frac{24}{x-2}).

Tip: Synthetic division is faster for linear divisors of the form (x-c); just remember to change the sign of (c) when you write the synthetic row.

3. Common Mistakes and How to Avoid Them

4. Frequently Asked Questions (FAQ)

Q1: Can I use the quadratic formula for every quadratic equation?
A: Yes. The formula works for all quadratics, regardless of whether they factor nicely. On the flip side, factoring is quicker when possible, and completing the square provides insight for graphing.

Q2: When is it better to complete the square instead of using the formula?
A: Completing the square is ideal when you need the vertex form or when the coefficients make the discriminant a perfect square, which simplifies the arithmetic.

Q3: How do I know if a quadratic expression is a perfect square trinomial?
A: If the first and last terms are perfect squares (e.g., (x^{2}) and (9)) and the middle term is twice the product of their square roots (e.g., (2\cdot x \cdot 3 = 6x)), then it is a perfect square: ((x+3)^{2}).

Q4: Why does the discriminant determine the number of real solutions?
A: The discriminant is the value under the square root in the quadratic formula. A positive value yields two distinct real numbers, zero yields a single repeated real number, and a negative value yields an imaginary number because you cannot take the square root of a negative in the real number system.

Q5: In word problems, how can I quickly set up the quadratic equation?
A: Identify the variable (often time, distance, or length), translate each statement into an algebraic expression, and bring all terms to one side so the equation equals zero. Look for patterns like “area = length × width” or “height = -5t² + vt + h₀”.

5. Study Strategies for Mastering Unit 4

1. Create a “Formula Cheat Sheet.” Write the vertex form, standard form, quadratic formula, and discriminant condition on a single index card. Review it daily.
2. Practice Reverse Engineering. Take a graph, read off the vertex and intercepts, then write the equation in both vertex and standard forms. This reinforces the connection between algebraic and visual representations.
3. Use Real‑World Contexts. Convert a physics problem (projectile motion) or an economics problem (maximizing profit) into a quadratic model. The narrative makes abstract symbols more memorable.
4. Timed Drills. Set a timer for 5‑minute blocks and solve as many factoring problems as possible. Speed builds confidence for the quiz portion of the unit.
5. Teach a Peer. Explaining the “why” behind each step (e.g., why we add and subtract the same number when completing the square) reveals gaps in your own understanding.

6. Final Checklist Before Turning in Homework 3

• [ ] All answers are in simplest form (fractions reduced, radicals rationalized).
• [ ] Each step is clearly labeled (e.g., “Factor the quadratic”, “Apply quadratic formula”).
• [ ] Units are included where appropriate (seconds, meters, dollars).
• [ ] Graph sketches, if required, show vertex, axis of symmetry, and intercepts with correct scale.
• [ ] The final answer for each problem is boxed or highlighted for easy grading.

Conclusion

Unit 4 of All Things Algebra is a key point where students transition from linear reasoning to the richer, more visual world of quadratics. Here's the thing — by following the step‑by‑step solutions, internalizing the common pitfalls, and employing the study strategies outlined above, you will not only complete the assignment with confidence but also lay a solid foundation for the more advanced topics that follow in later units. Homework 3 consolidates this transition by demanding fluency in multiple solution methods, an ability to interpret graphs, and the skill to translate real‑life scenarios into algebraic language. Keep practicing, stay organized, and remember that every quadratic you solve sharpens your problem‑solving intuition—an invaluable asset in mathematics and beyond.