Ten short lessons that teach the Unit 4 quadratics methods used in the Georgia DOE lesson plans, in the order those plans teach them. Each lesson ends with the test study-guide problems it prepares for. Everything follows the state's own lesson plans and teacher guides, so you won't introduce a method the class hasn't seen.
The class follows the Georgia DOE Algebra: Concepts & Connections Unit 4 lessons almost exactly. Those lessons build everything from pictures (area models made with algebra tiles) and from "what's the same quantity written two ways." They lean on vocabulary more than on shortcuts.
The best coaching move, straight from the teacher guides, is asking your student to explain. Each lesson ends with an Ask your student box. If they can explain it, they own it. If they can't, that's the spot to work on.
A quadratic graphs as a parabola. The same parabola can be written three ways, and the unit's big idea is that each form shows you a different feature for free. Rewriting from one form to another is how you find the feature you need.
| Form | Looks like | Shows you directly |
|---|---|---|
| Standard | y = ax2 + bx + c | a: opens up (a > 0) or down (a < 0). c: the y-intercept (the starting height in a word problem) |
| Vertex | y = a(x − h)2 + k | Vertex (h, k), axis of symmetry x = h. k is the maximum or minimum value. |
| Intercept | y = a(x − p)(x − q) | x-intercepts p and q. The vertex sits halfway between them. |
These are the same profit function (from the "Seeing Structure in Expressions" lesson). The intercept form shows the company breaks even (profit 0) at 2 and at 20 items. The vertex form shows the most profit is 81, at 11 items. Opening down means a maximum. And 11 sits exactly halfway between 2 and 20.
The questions this lesson prepares for, exactly as on the study guide. Each button loads that question into this lesson's explorer.
The opening of a tunnel through a mountain is modeled by the function h(x) = −130x2 + 2x, where h is the height of the tunnel (in feet) and x is the distance (in feet) from the bottom left corner of the tunnel opening. What is the width across the bottom of the tunnel opening?
The printed guide reads "−301 x²". The fraction 1/30 lost its slash.
Factor: x(−130x + 2) = 0, so x = 0 or x = 60. The tunnel is 60 ft wide.
The class learned vertex form by starting from the parent graph y = x2 and asking what each number does to it.
| In a(x − h)2 + k | Effect on y = x2 |
|---|---|
| + k | Shift up k (down if k is negative) |
| (x − h) | Shift right h. So (x − 2) is right 2 and (x + 2) is left 2. |
| a > 1 | Vertical stretch (looks narrower) |
| 0 < a < 1 | Vertical shrink (looks wider). The guide stresses this means the outputs got smaller. |
| a < 0 | Reflect over the x-axis (opens down) |
y = −(x + 2)2 − 5
The class treats a product as the area of a rectangle. One factor goes across the top and the other down the side. Each cell is a small area, and the answer is the total area with like terms combined. In class they first built these with algebra tiles.
Worked example, the first problem on the class packet: (x − 3)(3x + 1)
Cells: 4x2, 32x, −2x, −16. Total: 4x2 + 30x − 16.
The questions this lesson prepares for, exactly as on the study guide. Each button loads that question into this lesson's explorer.
Find the product.
a. 20x2 + 6x − 8 · b. 16x2 + 56x + 45
Factoring is the area model in reverse. You know the total area and you need the side lengths. The class mini-lesson taught pulling out the GCF first, so always check for one before anything else.
Worked example, x2 + 7x − 18. The corners multiply to −18x2. Which pair multiplies to that and adds to 7x? 9x and −2x.
When a ≠ 1, such as 2x2 + 5x − 3 from the class packet: the corners multiply to −6x2. The pair 6x and −x adds to 5x. The top row is 2x2, 6x, which share 2x. Working out the rest of the edges gives (2x − 1)(x + 3).
The GaDOE lesson teaches the box but doesn't spell out the "corners multiply" step. That step is the usual way to fill the box, and it's how DeltaMath's "Solve by Trinomial Factor (a>1)" problems work. If your student fills the box another way, such as guess-and-check with tiles, go with their way.
When the box comes out with both sides the same, you've made a square: x2 + 6x + 9 = (x + 3)2. To spot one, check that the first and last terms are perfect squares and the middle term is twice the product of their roots. This is the bridge to completing the square in Lesson 7.
The corners multiply to −210x2. The pair 21x and −10x adds to 11x. Answer: (3x − 5)(2x + 7).
The questions this lesson prepares for, exactly as on the study guide. Each button loads that question into this lesson's explorer.
Factor the polynomial.
a. (x − 4)2 · b. (3a + 1)2. Both are perfect square trinomials.
Find all values of b for which x2 + bx + 49 is a perfect square trinomial.
b = 14 or b = −14, from (x + 7)2 and (x − 7)2.
The questions this lesson prepares for, exactly as on the study guide. Each button loads that question into this lesson's explorer.
Write the equations in standard form:
a. −x2 − 23 = 0 · b. −5x2 − 9x + 3 = 0 · c. −7x2 − 2x + 6 = 0
Multiplying through by −1 gives an equally correct version with a positive leading term, such as 5x2 + 9x − 3 = 0. Use whichever the class prefers.
The class packet's own example, which is word for word study guide #6a: notice the left side is already a perfect square.
The questions this lesson prepares for, exactly as on the study guide. Each button loads that question into this lesson's explorer.
Solve each equation without graphing.
a. (x + 3)2 = 25, so 2, −8 · b. (x − 5)2 = 49, so 12, −2 · c. x2 = 25, so 5, −5
The class built this with algebra tiles. x2 + 6x is an x-by-x square plus six x-strips. Split the strips into two groups of 3, put them on two sides, and there's a missing corner of 3 × 3 = 9. Adding that 9 "completes the square."
Odd b gives fractions, which is where students stumble. For x2 − 3x: half of −3 is −32, and squared that's 94. Keep them as fractions rather than decimals.
The questions this lesson prepares for, exactly as on the study guide. Each button loads that question into this lesson's explorer.
Find the value of c that completes the square.
a. 64 · b. 14 · c. 254
Solve the equation by completing the square.
a. (x + 5)2 = 41, so x = −5 ± 41 ≈ 1.4, −11.4
b. (x − 32)2 = 374, so x = 3 ± 372 ≈ 4.5, −1.5
c. (x + 152)2 = 2734, so x = −15 ± 2732 ≈ 0.8, −15.8
A wading pool is 1 foot deep and has a volume of 108 cubic feet. The width is 12 feet less than the length.
a. l(l − 12)(1) = 108
b. Reject −6. The pool is 18 ft long, 6 ft wide (and 1 ft deep).
The product of two consecutive odd integers that are positive is 323. Write and solve an equation to find the integers.
The integers must be positive, so n = 17. They're 17 and 19.
The same "½ of b, squared" idea, but now you're rewriting a function rather than solving an equation. You can't add to both sides of a function, so the class packet says to add it and subtract it, which is adding zero.
This is the packet's own example, and it's the step students miss most. Whatever you add inside the parentheses gets multiplied by a.
The questions this lesson prepares for, exactly as on the study guide. Each button loads that question into this lesson's explorer.
The function I(t) = −0.2t2 + 3.4 represents the yearly income or loss I (in dollars) from a real estate investment, where t is the time (in years) after 2010. During what year did the investment earn the maximum amount of income?
As printed, this is trivially t = 0. It may have been meant as + 3.4t. Check with the teacher whether it was corrected in class.
As printed: the vertex is (0, 3.4), so the maximum is in 2010.
If it's −0.2t2 + 3.4t: completing the square gives −0.2(t − 8.5)2 + 14.45, so t = 8.5, during 2018.
The cost C (in dollars) of producing car tires in a factory is C(x) = 0.00003x2 − 0.06x + 65, where x represents the number of tires produced.
The printed guide reads "0.00003 − 0.06x", which is missing the x2.
a. 1,000 tires · b. $35 each
Determine whether the quadratic function has a maximum or minimum value. Then find the value.
a. (x − 3)2 − 5: minimum, −5
b. −3(x + 52)2 − 94: maximum, −9/4 (−2.25)
c. 2(x + 4)2 − 39: minimum, −39
d. 5(x − 2)2 + 5: minimum, 5
Write the quadratic function in vertex form. Identify the vertex and graph the function.
a. (x − 1)2 + 5, vertex (1, 5) · b. (x + 3)2 − 8, vertex (−3, −8)
c. −(x + 1)2 − 4, vertex (−1, −4), opens down · d. 3(x − 1)2 − 1, vertex (1, −1)
To graph: plot the vertex, then use "over 1, over 2" from Lesson 2.
The class meets this one last, as a tool for "pesky equations" that are painful to factor or complete. The teacher guide says it comes from completing the square on ax2 + bx + c = 0, but students aren't asked to derive it.
The guide's demo: x2 − 8x + 15 = 0, so a = 1, b = −8, c = 15.
Study guide #1 and #16 use the falling-object model h = −16t2 + vt + h0, where v is the launch speed and h0 is the starting height. "When is it at 9 feet?" means: set the formula equal to 9, get standard form = 0, then use the quadratic formula. Keep only times that make sense.
The questions this lesson prepares for, exactly as on the study guide. Each button loads that question into this lesson's explorer.
The height h (in feet) of a fly ball in a baseball game can be modeled by h = −16t2 + 28t + 16, where t is the time (in seconds).
The printed guide reads "−162", which is missing the t.
a. The t-intercepts are the times when the height is 0. The positive one is when the ball lands. The negative one is before the hit, so it has no meaning here.
b. Divide by −4 to get 4t2 − 7t − 4 = 0. The quadratic formula gives t = 7 ± 1138, so about 2.2 seconds. Reject −0.45.
You launch a water balloon. The function h = −16t2 + 45t + 2 models the height h (in feet) of the water balloon after t seconds.
The printed guide reads "−162", which is missing the t.
a. −16t2 + 45t − 7 = 0, so t = 45 ± 157732 ≈ 0.2 s and 2.6 s. Both make sense: once going up, once coming down.
b. −16t2 + 45t + 2 = 0, so about 2.9 s. Reject the negative root.
Solve the equation using the Quadratic Formula. Round your solutions to the nearest tenth, if necessary.
a. 4 (the discriminant is 0) · b. 1, −11 · c. −9 ± 17716 ≈ 0.3, −1.4 · d. The discriminant is −76, so no real solutions.
For c and d, rewrite in standard form = 0 before reading off a, b and c.
The test will ask students to pick a method and say why. The class's reasoning, summarized from the teacher guide:
| If the equation… | Use | Example |
|---|---|---|
| has no x-term | Square root method | x2 − 1 = 24 |
| has no constant | Factor out the GCF, then zero-product | −7x2 = 21x |
| factors easily | Factor, then zero-product | x2 + 10x − 11 = 0 |
| has a = 1 and an even b | Completing the square | x2 − 20x = 13 |
| is messy, or a ≠ 1 | Quadratic formula (it always works) | 8x2 + 9x = 3 |
The questions this lesson prepares for, exactly as on the study guide. Each button loads that question into this lesson's explorer.
Solve the equation using any method. Explain your choice of method.
a. a = 1 and b is even, so complete the square: (x − 10)2 = 113, giving x = 10 ± 113 ≈ 20.6, −0.6.
b. There's no constant, so move everything to one side and factor out the GCF: −7x(x + 3) = 0, giving 0 and −3. Don't divide by x, or you lose 0.
Every question now sits inside the lesson that teaches it, with its answer and a button that loads it into that lesson's explorer. Use this table to jump to any one.
#1, #3, #5 and #16 lost a t, an x2 or a fraction slash. Their cards show the fixed version and what the printed copy reads. #4 is genuinely unclear; check with the teacher whether it was corrected in class.
| # | What it asks | Where it's taught |
|---|---|---|
| #1 | Fly ball: t-intercepts and landing time corrected | 9 · Quadratic formula |
| #2 | Write equations in standard form | 5 · Zero-product |
| #3 | Tunnel width corrected | 1 · Three forms |
| #4 | Year of maximum income unclear | 8 · Vertex form |
| #5 | Tire cost minimum corrected | 8 · Vertex form |
| #6 | Solve without graphing (perfect squares) | 6 · Square roots |
| #7 | Multiply binomials | 3 · Multiplying |
| #8 | Factor perfect squares | 4 · Factoring |
| #9 | Value of c that completes the square | 7 · Completing the square |
| #10 | Solve by completing the square | 7 · Completing the square |
| #11 | Wading pool dimensions | 7 · Completing the square |
| #12 | Values of b for a perfect square | 4 · Factoring |
| #13 | Consecutive odd integers, product 323 | 7 · Completing the square |
| #14 | Maximum or minimum value | 8 · Vertex form |
| #15 | Vertex form, vertex and graph | 8 · Vertex form |
| #16 | Water balloon: at 9 ft, and landing corrected | 9 · Quadratic formula |
| #17 | Quadratic formula | 9 · Quadratic formula |
| #18 | Any method, and explain why | 10 · Choosing a method |
Factoring when a > 1, completing the square and vertex form are where students most often slip, and the quadratic formula comes last in the unit. Spend the most time on Lessons 4, 7, 8 and 9.