Gina Wilson All Things Algebra Unit 5 Homework 6

Unit 5 of Gina Wilson’s All Things Algebra centers on quadratic functions and their applications. This unit builds on students’ prior knowledge of linear equations and introduces the parabolic shape, factoring techniques, and the vertex form of a quadratic. Homework 6 specifically asks learners to solve a set of quadratic equations, graph corresponding parabolas, and interpret real‑world scenarios modeled by quadratics. The following article walks through the essential concepts, provides a detailed solution pathway, highlights frequent errors, and offers practical strategies for mastering the material.

Overview of Unit 5 Content

Key Topics Covered

• Standard form of a quadratic function: $ax^{2}+bx+c=0$.
• Factoring quadratics when possible.
• Completing the square to derive vertex form.
• Quadratic formula as a universal solving method.
• Graphical interpretation: axis of symmetry, vertex, direction of opening, and intercepts.
• Application problems involving area, projectile motion, and optimization.

Learning Objectives

1. Accurately factor quadratic expressions.
2. Apply the quadratic formula to find real and complex roots.
3. Convert between standard and vertex forms.
4. Sketch precise graphs of quadratic functions.
5. Translate word problems into quadratic equations and solve them.

Understanding Homework 6

Homework 6 typically consists of six distinct problems that reinforce the above objectives. While the exact wording may vary by edition, the structure generally includes:

1. Factoring exercises – rewriting quadratics as products of binomials.
2. Formula application – solving equations that do not factor neatly.
3. Graphing tasks – plotting vertex, axis of symmetry, and intercepts.
4. Word‑problem translation – setting up and solving real‑world quadratic models.

Each problem requires a clear, step‑by‑step response, making it essential to understand both the procedural mechanics and the underlying concepts The details matter here..

Step‑by‑Step Solution Guide

Problem 1: Factoring a Simple Quadratic

Solve: $x^{2}-5x+6=0$.

Solution Process

1. Identify two numbers that multiply to $6$ (the constant term) and add to $-5$ (the coefficient of $x$).
   The numbers are $-2$ and $-3$.
2. Write the quadratic as $(x-2)(x-3)=0$.
3. Set each factor equal to zero: $x-2=0$ or $x-3=0$.
4. Solve: $x=2$ or $x=3$.

Result: The solutions are $x=2$ and $x=3$ Nothing fancy..

Problem 2: Using the Quadratic Formula

Solve: $2x^{2}+3x-2=0$.

Solution Process

1. Identify coefficients: $a=2$, $b=3$, $c=-2$.
2. Compute the discriminant: $\Delta = b^{2}-4ac = 3^{2}-4(2)(-2)=9+16=25$.
3. Apply the quadratic formula:
   $x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-3\pm5}{4}$.
4. Calculate the two roots:
   $x_{1}=\frac{-3+5}{4}= \frac{2}{4}=0.5$,
   $x_{2}=\frac{-3-5}{4}= \frac{-8}{4}= -2$.

Result: $x=0.5$ and $x=-2$.

Problem 3: Completing the Square

Rewrite $x^{2}+6x+5$ in vertex form.

Solution Process

1. Take half of the coefficient of $x$: $\frac{6}{2}=3$.
2. Square it: $3^{2}=9$.
3. Add and subtract this square inside the expression:
   $x^{2}+6x+9-9+5 = (x+3)^{2}-4$.
4. The vertex form is $(x+3)^{2}-4$, indicating a vertex at $(-3,-4)$.

Result: Vertex form: $(x+3)^{2}-4$.

Problem 4: Graphing a Quadratic

Graph $y = -x^{2}+4x-3$.

Solution Process

1. Determine the direction of opening: coefficient of $x^{2}$ is negative → opens downward.
2. Find the vertex using $x = -\frac{b}{2a}$:
   $x = -\frac{4}{2(-1)} = 2$.
   Substitute to find $y$: $y = -(2)^{2}+4(2)-3 = -4+8-3 = 1$.
   Vertex: $(2,1)$.
3. Compute the axis of symmetry: $x=2$.
4. Find the $y$‑intercept: set $x=0$, $y=-3$.
5. Plot additional points (e.g., $x=1$ gives $y=0$, $x=3$ gives $y=0$) and draw a smooth parabola.

Result: A downward‑opening parabola with vertex $(2,1)$, axis $x=2$, and intercepts at $(0,-3)$, $(1,0)$, $(3,0)$ Still holds up..

Problem 5: Real‑World Application – Projectile Motion

A ball is launched upward with an initial velocity of 20 m/s from a height of 5 m. Its height (in meters) after $t$ seconds is given by $h(t)= -5t^{2}+20t+5$. When does the ball hit the ground?

Solution Process

1. Set $h(t)=0$: $-5t^{2}+20t+5=0$.
2. Multiply by $-1$ to simplify: $5t^{2}-20t-5=0$.
3. Apply the quadratic formula with $a=5$, $b=-20$, $c=-5$:
   $\Delta = (-20)^{2}-4(5)(-5)=400+100=500$.
4. $t=\frac{20\pm\sqrt{500}}{10}= \frac{20\pm10\sqrt{5}}{10}=2\pm\sqrt{5}$.
5. Since time cannot be negative, take the positive root: $t=2+\sqrt{5}\approx4.24$ seconds.

Result: The ball reaches the ground after approximately $4.24$ seconds The details matter here. No workaround needed..

Problem 6: Optimization – Maximizing Area

A rectangular garden is to be enclosed on three sides by fencing, with the fourth side being a river. If 120 m of fencing is available, what dimensions maximize the area?

Solution Process

1. Let $x$ be the length

perpendicular to the river and (y) be the length parallel to the river.

2. Since only three sides need fencing:
   [ 2x+y=120 ] Solve for (y):
   [ y=120-2x ]

3. Write the area formula:
   [ A=xy=x(120-2x)=120x-2x^{2} ]

4. This is a downward-opening quadratic, so the maximum area occurs at the vertex:
   [ x=-\frac{b}{2a}=-\frac{120}{2(-2)}=30 ]

5. Substitute (x=30) into (y=120-2x):
   [ y=120-2(30)=60 ]

Result: The garden should be (30) m wide perpendicular to the river and (60) m long parallel to the river. The maximum area is:
[ A=30 \times 60=1800\text{ m}^{2} ]

Conclusion

Quadratic equations and functions are powerful tools for solving a wide range of mathematical and real-world problems. Whether you are solving by factoring, using the quadratic formula, completing the square, graphing, modeling projectile motion, or optimizing area, the same key ideas apply: identify the structure of the quadratic, use the vertex when needed, and interpret your answer in context Small thing, real impact..

Mastering these methods builds a strong foundation for algebra, physics, engineering, economics, and many other fields where quadratic relationships naturally appear.

Problem 7: Quadratic Inequalities – Safe Driving Distance

A driver must maintain a safe following distance that satisfies the inequality (d(t) = 3t^2 - 20t + 50 > 0), where (d) is the distance (in meters) after (t) seconds of acceleration. For how long is the driver safe to keep the car in the same lane?

Solution Process

1. Find the roots of the quadratic equation (3t^2 - 20t + 50 = 0) to locate the boundary points of the inequality.
   Using the quadratic formula:
   [ t = \frac{20 \pm \sqrt{(-20)^2-4(3)(50)}}{2(3)} = \frac{20 \pm \sqrt{400-600}}{6} ] The discriminant is negative ((-200)), so the quadratic never crosses the (t)-axis.
2. Since the coefficient of (t^2) is positive, the parabola opens upward and is always above zero for all real (t).
3. That's why, the inequality holds for every non‑negative time value.
4. In practical terms, the driver can safely maintain the lane indefinitely as long as no other constraints (speed limits, traffic lights, etc.) intervene.

Result: The inequality is satisfied for all (t \ge 0); the driver is always within a safe distance.

Problem 8: Quadratic Sequences – Pattern Recognition

Consider the sequence defined by (a_n = n^2 - 3n + 2). Identify the first five terms and determine whether the sequence is increasing, decreasing, or neither.

Solution Process

1. Compute the first five terms:

   • (a_1 = 1^2 - 3(1) + 2 = 0)
   • (a_2 = 4 - 6 + 2 = 0)
   • (a_3 = 9 - 9 + 2 = 2)
   • (a_4 = 16 - 12 + 2 = 6)
   • (a_5 = 25 - 15 + 2 = 12)

2. Observe the differences:

   • (0 \to 0) (Δ = 0)
   • (0 \to 2) (Δ = 2)
   • (2 \to 6) (Δ = 4)
   • (6 \to 12) (Δ = 6)

3. The differences themselves increase by 2 each step, indicating that the sequence accelerates upward The details matter here..

4. Since all subsequent differences are positive, the sequence is strictly increasing for (n \ge 3) Easy to understand, harder to ignore..

Result: The sequence begins (0, 0, 2, 6, 12) and increases monotonically after the second term.

Problem 9: Quadratic Recurrence – Long‑Term Behavior

Let (x_{n+1} = 2x_n^2 - 1) with an initial value (x_0 = \cos(\theta)). Show that (x_n = \cos(2^n \theta)) for all non‑negative integers (n).

Solution Process

1. Base case: For (n=0), (x_0 = \cos(\theta)) matches the formula.
2. Inductive step: Assume (x_n = \cos(2^n \theta)).
   Then [ x_{n+1} = 2x_n^2 - 1 = 2\cos^2(2^n \theta) - 1 = \cos(2 \cdot 2^n \theta) = \cos(2^{n+1}\theta). ]
3. By induction, the identity holds for all (n).

Result: The recurrence produces successive double‑angle cosines, a classic example of a quadratic map generating periodic or chaotic behavior depending on (\theta).

Final Thoughts

Quadratics permeate both abstract mathematics and everyday life. Whether you’re sketching a parabola, calculating projectile trajectories, optimizing resources, or exploring iterative maps, the core techniques—factoring, the quadratic formula, completing the square, and vertex analysis—remain indispensable. By mastering these tools, you gain a versatile framework that extends to higher‑degree polynomials, differential equations, and even modern computational algorithms.

Feel encouraged to experiment further: try fitting quadratic models to real data, investigate the effects of parameter changes on the shape of a parabola, or explore how quadratic inequalities define feasible regions in optimization problems. The more you practice, the more intuitive these concepts become, and the more powerful your mathematical toolkit will be Took long enough..