8.11Unit 8 Review — Quadratic Functions
A complete review of every Unit 8 concept — graphing parabolas, vertex form, factoring, the quadratic formula, the discriminant, solving by square roots, and quadratic applications — with worked examples, mixed practice, and a final checklist to prepare for your unit test.
Why This Matters
Quadratic functions are one of the most tested topics on the SAT and ACT. Reviewing this unit thoroughly prepares you for Algebra 2, AP Physics kinematics, and the polynomial and function units in Precalculus.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Unit 8 Review
Quadratic Functions — Complete Review
Covers Chapters 01–10. Use this chapter to prepare for your Unit 8 Test.
Unit 8 Concept Map
Unit 8 Concept Map
Standard Form
ax²+bx+c
Vertex Form
a(x−h)²+k
Graphing
Parabola, vertex, axis
Solving
Factor / √ / CTS / Formula
Discriminant
b²−4ac → # solutions
Applications
Projectile, area, revenue
Key Vocabulary Review
Quadratic Function
A function of the form f(x)=ax²+bx+c where a≠0. Its graph is a parabola.
Parabola
The U-shaped graph of a quadratic function. Opens up when a>0, down when a<0.
Vertex
The highest or lowest point of a parabola. Coordinates: (h, k) or (−b/2a, f(−b/2a)).
Axis of Symmetry
The vertical line x = −b/(2a) that passes through the vertex and divides the parabola into mirror halves.
Standard Form
y = ax²+bx+c. Reveals y-intercept (0,c) and coefficients for formula use.
Vertex Form
y = a(x−h)²+k. Reveals vertex (h,k) and transformations directly.
x-Intercept / Root / Zero
Where the parabola crosses the x-axis. Found by setting y=0 and solving.
Discriminant
D = b²−4ac. Positive → 2 solutions; zero → 1 solution; negative → no real solutions.
Square Root Property
If x²=k, then x=±√k. Efficient when no bx term is present.
Zero Product Property
If A·B=0, then A=0 or B=0. The basis for solving by factoring.
Quadratic Formula
x=(−b±√(b²−4ac))/(2a). Solves any quadratic equation in standard form.
Quadratic Model
A quadratic function used to represent a real-world situation such as projectile motion, area, or revenue.
Formula Reference Card
Unit 8 Formula Reference Card
Topic 1 — Graphing Quadratic Functions
Every quadratic function in standard form y=ax²+bx+c produces a parabola. The sign of a controls opening direction; the vertex is the extreme point; the axis of symmetry is the vertical line through the vertex. Use the checklist below to graph any parabola systematically.
Parabola Anatomy — y = 0.4(x−1)²−2
Vertex (1, −2)
Minimum point. a > 0 → opens up.
Axis of Symmetry x=1
Vertical line through the vertex. x = −b/(2a).
x-Intercepts
Where y=0. Found by factoring or quadratic formula.
Parabola
U-shaped curve. Opens up when a>0, down when a<0.
Opening Direction — Maximum vs. Minimum
a > 0 — Opens Up
Vertex is the minimum. Parabola has a lowest point.
a < 0 — Opens Down
Vertex is the maximum. Parabola has a highest point.
Graphing a Quadratic — Step-by-Step Checklist
Topic 2 — Vertex Form & Transformations
Vertex form y=a(x−h)²+k makes the vertex (h,k) and all transformations immediately visible. Remember: the sign inside the parentheses is opposite — (x−3)² has h=+3, not −3.
Standard Form vs. Vertex Form
Standard Form
y = ax² + bx + c
y-intercept: (0, c) — read directly
Axis of symmetry: x = −b/(2a)
Vertex: plug x = −b/(2a) back in
Opens up: a > 0 | Opens down: a < 0
Best for: finding y-intercept, solving by factoring/formula
Vertex Form
y = a(x − h)² + k
Vertex: (h, k) — read directly
Axis of symmetry: x = h
Max/Min: k is the max (a<0) or min (a>0)
Opens up: a > 0 | Opens down: a < 0
Best for: graphing, transformations, max/min problems
Vertex Form Transformations — y = a(x−h)²+k vs. y = x²
| Parameter | Effect on graph | Example |
|---|---|---|
| h > 0 | Shift right h units | y=(x−3)² → right 3 |
| h < 0 | Shift left |h| units | y=(x+2)² → left 2 |
| k > 0 | Shift up k units | y=x²+5 → up 5 |
| k < 0 | Shift down |k| units | y=x²−4 → down 4 |
| |a| > 1 | Vertical stretch (narrower) | y=3x² → narrower |
| 0 < |a| < 1 | Vertical compression (wider) | y=0.5x² → wider |
| a < 0 | Reflection over x-axis (opens down) | y=−x² → flips |
Two Parabolas — y=x²−4 (blue) vs. y=−0.5x²+3 (purple)
Topic 3 — Solving by Factoring
To solve by factoring: write in standard form, factor completely, set each factor equal to zero, and solve. Never divide both sides by a variable — you lose solutions.
Factoring Methods — Quick Summary
Topic 4 — The Quadratic Formula & Discriminant
The quadratic formula x=(−b±√(b²−4ac))/(2a) works on every quadratic. Always check the discriminant first — it tells you how many solutions to expect before you do any arithmetic.
Discriminant Quick Reference — D = b²−4ac
D > 0
2 real solutions
Crosses x-axis twice
x²−5x+4=0, D=9
D = 0
1 real solution (double root)
Touches x-axis once
x²−6x+9=0, D=0
D < 0
No real solutions
Does not cross x-axis
x²+x+1=0, D=−3
Topic 5 — Solving by Square Roots
The square root method is the most efficient approach when the equation has no bx term — that is, when it is in the form ax²=k or a(x−h)²=k. Isolate the squared expression, apply ±√, and solve for x. If the value under the radical is negative, there are no real solutions.
Square Root Property — Quick Reference
Basic form
x² = k → x = ±√k
Example: x²=49 → x=±7
Vertex-style form
(x−h)² = k → x = h ± √k
Example: (x−3)²=16 → x=3±4
With coefficient
a(x−h)² = k → (x−h)² = k/a
Example: 2(x+1)²=18 → (x+1)²=9 → x=−1±3
No real solution
x² = −k (k> 0) → no real solution
Example: x²=−9 → no real solution
Topic 6 — Quadratic Applications and Modeling
Quadratic models appear in projectile motion, area optimization, and revenue problems. Identify what the question asks — maximum/minimum (vertex), when output is zero (zeros), starting value (y-intercept) — and choose the appropriate feature. Always interpret results using units and domain restrictions.
Context-to-Feature Organizer
"Maximum height / minimum cost"
→ Vertex (h, k)
"When does it hit the ground?"
→ Zeros (x-intercepts)
"Starting value / initial height"
→ y-intercept (0, c)
"Height at t = 3 seconds"
→ Evaluate f(3)
Choosing a Solving Method
Solving Method Decision Flowchart
Method
Fast when no bx term
quickly?
Zero Product
Formula
Common Mistakes — All Topics
Unit 8 — Top 10 Common Mistakes
Topic: Axis of symmetry
✗ x = b/(2a) (forgot the negative)
✓ x = −b/(2a)
Topic: Vertex form sign
✗ y=(x+3)²+1 → vertex (3,1)
✓ y=(x+3)²+1 → vertex (−3,1) because h=−3
Topic: Opening direction
✗ a=−2 → opens up
✓ a<0 → opens DOWN
Topic: Factoring signs
✗ x²−5x+6=(x+2)(x+3)
✓ (x−2)(x−3): both negative, product=+6, sum=−5
Topic: Dividing by x
✗ x²+3x=0 → divide by x → x=−3 only
✓ Factor: x(x+3)=0 → x=0 or x=−3
Topic: Quadratic formula −b
✗ x²+4x+3=0 → x=(4±√4)/2
✓ x=(−4±√4)/2 = (−4±2)/2 → x=−1 or x=−3
Topic: Denominator 2a
✗ 2x²+5x+2=0 → x=(−5±3)/2
✓ Denominator is 2a=4: x=(−5±3)/4 → x=−1/2 or x=−2
Topic: Square root ±
✗ x²=25 → x=5 only
✓ x=±5; both +5 and −5 satisfy x²=25
Topic: Negative discriminant
✗ D=−4 → √(−4)=2i → x=... (gives complex answer)
✓ D<0 → no real solutions. Stop.
Topic: Context domain
✗ Projectile: t=−0.5 s is a valid answer
✓ Reject negative time; only t≥0 is meaningful in context.
Practice Grids
Practice Grid — Use for Graphing Problems
Worked Review Examples
Graph y = x² − 4x + 3. Find vertex, axis, intercepts.
a=1, b=−4, c=3. Opens up (a>0).
Axis: x=−(−4)/(2·1)=2
Vertex: y=(2)²−4(2)+3=4−8+3=−1 → vertex (2,−1)
y-intercept: (0,3)
x-intercepts: x²−4x+3=0 → (x−1)(x−3)=0 → x=1 or x=3
Write y = x² + 6x + 5 in vertex form.
Axis: x=−6/2=−3
Vertex y: (−3)²+6(−3)+5=9−18+5=−4 → vertex (−3,−4)
Vertex form: y=(x+3)²−4
Solve x² − 2x − 15 = 0 by factoring.
Find p,q: pq=−15, p+q=−2 → p=−5, q=3
(x−5)(x+3)=0
x=5 or x=−3
Use the discriminant to classify solutions: 4x² − 4x + 1 = 0
D = (−4)²−4(4)(1) = 16−16 = 0
D=0 → exactly one real solution (double root)
x = 4/(2·4) = 1/2
Solve 3(x − 2)² = 75 using the square root method.
Divide both sides by 3: (x−2)²=25
Take square root: x−2=±5
x=2+5=7 or x=2−5=−3
Check: 3(7−2)²=3(25)=75 ✓ and 3(−3−2)²=3(25)=75 ✓
A ball is thrown upward: h(t) = −16t² + 64t + 6. Find the maximum height and when it hits the ground.
Maximum height at vertex: t = −64/(2·(−16)) = 64/32 = 2 s
h(2) = −16(4)+64(2)+6 = −64+128+6 = 70 ft
Hits ground when h=0: −16t²+64t+6=0 → 8t²−32t−3=0
D=1024+96=1120; t=(32+√1120)/16 ≈ (32+33.47)/16 ≈ 4.09 s (reject negative)
Common Mistakes
Confusing the vertex formula — using x = b/2a instead of x = −b/(2a).
The axis of symmetry is x = −b/(2a). The negative sign is critical.
Forgetting ± when applying the square root property: x²=25 → x=5 only.
x²=25 → x=±5. Both solutions must be included.
Setting each factor equal to zero but forgetting to solve — stopping at (x + 3)(x − 2) = 0.
Set each factor equal to zero and solve: x + 3 = 0 → x = −3; x − 2 = 0 → x = 2.
Reporting a negative time or length as a valid answer in a context problem.
Check domain. Reject solutions that are not physically meaningful (e.g., negative time).
Mixed Review Practice — 22 Problems
Guided Practice Video: Unit 8 Review
Watch this full Unit 8 review covering quadratic functions, vertex form, factoring, the quadratic formula, discriminant, completing the square, complex numbers, and applications before working through the mixed practice below.
Video by Sang Real Math
Watch on YouTube ↗Find the axis of symmetry and vertex of y = x² − 6x + 8.
Graph y = −x² + 4. State vertex, opening direction, and x-intercepts.
Write y = x² − 8x + 7 in vertex form.
Identify all transformations of y = 3(x + 2)² − 5.
Solve by factoring: x² + 7x + 12 = 0.
Solve by factoring: 2x² − 3x − 2 = 0.
Solve using the square root property: x² = 81.
Solve using the square root property: (x + 4)² = 49.
Solve: 5(x − 1)² = 80.
Use the discriminant to classify: x² + 5x + 7 = 0.
Use the discriminant to classify: x² − 4x − 5 = 0.
Solve using the quadratic formula: x² + 4x − 1 = 0. Leave in exact form.
Solve using the quadratic formula: 2x² − 6x + 3 = 0.
A ball is thrown: h = −16t² + 48t. Find the maximum height.
The area of a rectangle is 28. Length is (x+3), width is (x−1). Find x.
Convert y = (x − 4)² − 9 to standard form.
Find the x-intercepts of y = x² + 2x − 8.
For y = −2x² + 8x − 3, find the maximum value.
A farmer has 80 m of fencing for 3 sides of a rectangle. Find the dimensions that maximize area.
Error Analysis: A student says the vertex of y=(x+5)²−3 is (5,−3). Correct the error.
Solve x² − 5x = 0 and explain why you cannot divide both sides by x.
Explain in one sentence why the square root method is not efficient for x²+5x−6=0.
Challenge Problems
Find all values of c such that x²+6x+c=0 has two distinct real solutions.
A parabola has vertex (2,−3) and passes through (4,5). Write its equation in vertex form and standard form.
Solve: x⁴−13x²+36=0 using substitution u=x².
The sum of two numbers is 12 and their product is 35. Find the numbers using a quadratic equation.
A rectangle has perimeter 36 and area 80. Write and solve a quadratic equation for the dimensions.
A ball is thrown from a cliff 100 ft high with initial velocity 48 ft/s: h=−16t²+48t+100. Find the maximum height and when it hits the ground.
For what values of k does kx²+4x+1=0 have no real solutions?
Two parabolas y=x²−4 and y=−x²+4 intersect. Find the intersection points.
A quadratic has roots x=3+√2 and x=3−√2. Write the equation in standard form.
A company models profit as P(x)=−2x²+120x−800 where x is units sold. Find the break-even points and the maximum profit.
Final Review Checklist
Unit 8 Final Review Checklist