Solving Quadratic Equations Graphing Activity
Solving Quadratic Equations with Graphing
Name:
Student Mission (Read & Learn Independently): Follow this guide step-by-step. Each section introduces a concept with a completed model example followed by practice problems. Complete all steps directly on these sheets.
Part 1
Standard Form & Setting Equations Equal to Zero
A quadratic equation is in standard form when written as: ax² + bx + c = 0
Type A: Scrambled 1-Variable Equation \( x^2 - 5x = 14 \implies \) subtract 14: x² - 5x - 14 = 0
a = 1, b = -5, c = -14
Type B: Equation with y (Set y = 0) \( y = 2x^2 + 8x - 10 \implies \) set \( y = 0 \): 2x² + 8x - 10 = 0
a = 2, b = 8, c = -10
Rearrange each equation into standard form set to zero (\( ax^2 + bx + c = 0 \)), then identify \( a, b, \) and \( c \):
| Original Equation |
Standard Form (\( = 0 \)) |
Values of \( a, b, c \) |
| 1. \( x^2 + 6x = 7 \) |
|
a=__ b=__ c=__ |
| 2. \( y = x^2 - 9x + 18 \) |
|
a=__ b=__ c=__ |
| 3. \( 2x^2 = 8x - 6 \) |
|
a=__ b=__ c=__ |
| 4. \( y = -x^2 + 4x \) |
|
a=__ b=__ c=__ |
| 5. \( 5x - 14 = -x^2 \) |
|
a=__ b=__ c=__ |
Part 2
Identifying Roots from a Graph
Key Idea: Roots / Solutions are the \( x \)-values where the parabola crosses or touches the \( x \)-axis (\( y = 0 \)).
Model Example \( y = -x^2 + 4 \)
x y -2 2
Crosses \( x \): x = -2 & 2
Roots / sol.: x = -2, 2
Vertex: (0, 4) (Max)
Graph A \( y = x^2 - 4 \)
x y -2 2
Crosses \( x \):x = ___ & ___
Roots / sol.:x = _______
Vertex:(___, ___)
Graph B \( y = -(x-1)^2 + 4 \)
x y -1 3
Crosses \( x \):x = ___ & ___
Roots / sol.:x = _______
Vertex:(___, ___)
Technology Connection: Graphing with a Calculator or Desmos
Graph each equation using your graphing calculator or Desmos. Inspect your screen to identify the roots (\(x\)-intercepts) and vertex:
Problem C: y = x² - 6x + 5
Crosses \( x \)-axis at:x = _____ and _____
Roots / solutions:x = _____
Vertex coordinates:(___, ___)
Vertex type:min / max
Problem D: y = -x² - 2x + 8
Crosses \( x \)-axis at:x = _____ and _____
Roots / solutions:x = _____
Vertex coordinates:(___, ___)
Vertex type:min / max
Part 3
Identifying Roots & Vertices from a Table
The Table Secret: Since roots are points where \( y = 0 \), look for outputs where y = 0. The vertex is the symmetric turning point located right between them!
Model Example \( y = -x^2 + 1 \)
| x |
-2 |
-1 |
0 |
1 |
2 |
| y |
-3 |
0 |
1 |
0 |
-3 |
| Key Features: |
|
|
|
|
|
| ▲ Root (x = -1) |
|
|
|
|
|
|
▲
Vertex (0, 1)
Turning Point • Max
|
▲ Root (x = 1)
| |
Draw circular rings around the roots (\(y=0\)) in each table, then identify the vertex and whether it is a minimum or maximum:
Practice Table 1: \( y = x^2 - 2x - 3 \)
| x |
-2 |
-1 |
0 |
1 |
2 |
3 |
4 |
| y |
5 |
0 |
-3 |
-4 |
-3 |
0 |
5 |
a. Roots (where \(y=0\)): x = _____ & x = _____
b. Vertex (turning point): (___, ___)
c. Vertex type: min / max (circle one)
Practice Table 2: \( y = -x^2 + 4x \)
| x |
-1 |
0 |
1 |
2 |
3 |
4 |
5 |
| y |
-5 |
0 |
3 |
4 |
3 |
0 |
-5 |
a. Roots (where \(y=0\)): x = _____ & x = _____
b. Vertex (turning point): (___, ___)
c. Vertex type: min / max (circle one)
Part 4
Determining the Number of Real Solutions
Depending on where the parabola sits relative to the \( x \)-axis, it can have 2, 1, or 0 Real Solutions:
Case 1
2 Real Solutions
Crosses \( x \)-axis at two points.
Two \( x \)-intercepts
Case 2
1 Real Solution
Vertex touches at one point (\(x \neq 0\)).
10
One \( x \)-intercept (at \( x = 10 \))
Case 3
0 Real Solutions
Curve never touches \( x \)-axis.
No \( x \)-intercepts
Quick Check: State whether each situation has 0, 1, or 2 Real Solutions:
1. A parabola with vertex at \( (3, 0) \) ______ sol.
2. A parabola opening upward with vertex at \( (0, 5) \) ______ sol.
3. A parabola that crosses at \( x = -3 \) and \( x = 7 \) ______ sol.
4. A table where \( y \) is always negative numbers ______ sol.
Heads Up: What About Non-Real Solutions? When a graph has 0 real solutions, does that mean the equation has NO solutions at all? In Algebra 2, you will discover that equations like \( x^2 = -4 \) actually have two complex / imaginary solutions involving \( i = \sqrt{-1} \). We are not learning how to calculate those yet—so for right now, on our real graph, we state that there are 0 Real Solutions!
Part 5
Word Problems: Deciding Between Roots & Vertex
Translator Guide: What is the question asking you to find?
Finding the ROOTS (\( y = 0 \)):
Asked when an object hits the ground, lands, enters/exits water, or when a business breaks even (\(\text{profit} = 0\)).
Finding the VERTEX (\( h, k \)):
Asked for maximum height, highest peak, turnaround point, or maximum profit. Located halfway between the roots!
Scenario 1
The Oceanarium Acrobatic Dolphin Leap
Projectile Modeling
During an oceanarium show, a marine biologist models the leap of a dolphin using the quadratic equation: h(x) = -x² + 8x - 7, where \( h(x) \) is the dolphin's height in feet above the water surface and \( x \) is the horizontal distance traveled in feet.
Step A • Set Up Standard Form for Water Level
The water surface is at height \( h(x) = 0 \). Write the equation in standard form set to zero:
Step B • Find the Roots from the Flight Table
Use the provided height table below to identify the two roots where \( h(x) = 0 \):
| Distance x (ft) |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
| Height h (ft) |
-7 |
0 |
5 |
8 |
9 |
8 |
5 |
0 |
-7 |
• Dolphin emerges from water at: x = _____ ft
• Dolphin re-enters water at: x = _____ ft
Step C • Determine the Vertex (Peak of the Leap)
1. Halfway between the roots (\( x = 1 \) and \( x = 7 \)), what is the horizontal distance? x = _____ ft
2. What is the dolphin's maximum height at this vertex point? Height = _____ ft
3. Write the vertex as an ordered pair \( (x, y) \): (____, ____)
Step D • Solve the Trainer's Challenge Question
The trainer suspends a target ring 10 feet above the water surface directly above the middle of the leap. Does the dolphin reach the target ring? Explain using your calculations from Step C:
Scenario 2
Robotics Club Popcorn Fundraiser
Revenue & Profit
The high school robotics team sells gourmet popcorn tins to fund their competition travel. Their business advisor models their weekly profit using the quadratic equation: P(x) = -2x² + 28x - 80, where \( x \) represents the selling price per tin (in dollars) and \( P(x) \) represents the weekly profit (in hundreds of dollars).
Step A • Break-Even Equation Setup
"Breaking even" means the club makes $0 in profit (\( P(x) = 0 \)). Write the quadratic equation that must be solved to find the break-even selling prices:
Step B • Identify the Break-Even Roots
Examine the profit table computed for various selling prices:
| Selling Price x ($) |
$3 |
$4 |
$5 |
$6 |
$7 |
$8 |
$9 |
$10 |
$11 |
| Profit P ($100s) |
-14 |
0 |
10 |
16 |
18 |
16 |
10 |
0 |
-14 |
• Lower break-even price (\(P=0\)): $_____
• Upper break-even price (\(P=0\)): $_____
Step C • Determine the Maximum Profit (Vertex)
1. Halfway between the break-even prices ($4 and $10), what selling price maximizes profit? Price = $_____
2. What is the maximum profit generated at this optimal price? (Remember \( P \) is in hundreds of dollars): $_________
3. State the vertex coordinates: (_____, _____)
Step D • Business Evaluation Question
A student suggests pricing the popcorn tins at $15 each to make more money. Based on your knowledge of quadratic graphs and the roots, explain why pricing at $15 would actually cause the club to lose money:
Quadratic Graphing Answer Key
Answer Key
Solving Quadratic Equations with Graphing
Name: TEACHER & SUB KEY
Teacher & Substitute Guide: All solutions are printed in red slab-serif font for quick visual grading and step-by-step student verification.
Part 1
Standard Form & Setting Equations Equal to Zero
A quadratic equation is in standard form when written as: ax² + bx + c = 0
Type A: Scrambled 1-Variable Equation \( x^2 - 5x = 14 \implies \) subtract 14: x² - 5x - 14 = 0
a = 1, b = -5, c = -14
Type B: Equation with y (Set y = 0) \( y = 2x^2 + 8x - 10 \implies \) set \( y = 0 \): 2x² + 8x - 10 = 0
a = 2, b = 8, c = -10
Rearrange each equation into standard form set to zero (\( ax^2 + bx + c = 0 \)), then identify \( a, b, \) and \( c \):
| Original Equation |
Standard Form (\( = 0 \)) |
Values of \( a, b, c \) |
| 1. \( x^2 + 6x = 7 \) |
x² + 6x - 7 = 0 |
a=1, b=6, c=-7 |
| 2. \( y = x^2 - 9x + 18 \) |
x² - 9x + 18 = 0 |
a=1, b=-9, c=18 |
| 3. \( 2x^2 = 8x - 6 \) |
2x² - 8x + 6 = 0 |
a=2, b=-8, c=6 |
| 4. \( y = -x^2 + 4x \) |
-x² + 4x = 0 |
a=-1, b=4, c=0 |
| 5. \( 5x - 14 = -x^2 \) |
x² + 5x - 14 = 0 |
a=1, b=5, c=-14 |
Part 2
Identifying Roots from a Graph
Key Idea: Roots / Solutions are the \( x \)-values where the parabola crosses or touches the \( x \)-axis (\( y = 0 \)).
Model Example \( y = -x^2 + 4 \)
x y -2 2
Crosses \( x \): x = -2 & 2
Roots / sol.: x = -2, 2
Vertex: (0, 4) (Max)
Graph A \( y = x^2 - 4 \)
x y -2 2
Crosses \( x \): x = -2 & 2
Roots / sol.: x = -2, 2
Vertex: (0, -4) (Min)
Graph B \( y = -(x-1)^2 + 4 \)
x y -1 3
Crosses \( x \): x = -1 & 3
Roots / sol.: x = -1, 3
Vertex: (1, 4) (Max)
Technology Connection: Graphing with a Calculator or Desmos
Graph each equation using your graphing calculator or Desmos. Inspect your screen to identify the roots (\(x\)-intercepts) and vertex:
Problem C: y = x² - 6x + 5
Crosses \( x \)-axis at:x = 1 and 5
Roots / solutions:x = 1, 5
Vertex coordinates:(3, -4)
Vertex type:Minimum
Problem D: y = -x² - 2x + 8
Crosses \( x \)-axis at:x = -4 and 2
Roots / solutions:x = -4, 2
Vertex coordinates:(-1, 9)
Vertex type:Maximum
Part 3
Identifying Roots & Vertices from a Table
The Table Secret: Since roots are points where \( y = 0 \), look for outputs where y = 0. The vertex is the symmetric turning point located right between them!
Model Example \( y = -x^2 + 1 \)
| x |
-2 |
-1 |
0 |
1 |
2 |
| y |
-3 |
0 |
1 |
0 |
-3 |
| Key Features: |
|
|
|
|
|
| ▲ Root (x = -1) |
|
|
|
|
|
|
▲
Vertex (0, 1)
Turning Point • Max
|
▲ Root (x = 1)
| |
Draw circular rings around the roots (\(y=0\)) in each table, then identify the vertex and whether it is a minimum or maximum:
Practice Table 1: \( y = x^2 - 2x - 3 \)
| x |
-2 |
-1 |
0 |
1 |
2 |
3 |
4 |
| y |
5 |
0 |
-3 |
-4 |
-3 |
0 |
5 |
a. Roots (where \(y=0\)): x = -1 & x = 3
b. Vertex (turning point): (1, -4)
c. Vertex type: Minimum
Practice Table 2: \( y = -x^2 + 4x \)
| x |
-1 |
0 |
1 |
2 |
3 |
4 |
5 |
| y |
-5 |
0 |
3 |
4 |
3 |
0 |
-5 |
a. Roots (where \(y=0\)): x = 0 & x = 4
b. Vertex (turning point): (2, 4)
c. Vertex type: Maximum
Part 4
Determining the Number of Real Solutions
Depending on where the parabola sits relative to the \( x \)-axis, it can have 2, 1, or 0 Real Solutions:
Case 1
2 Real Solutions
Crosses \( x \)-axis at two points.
Two \( x \)-intercepts
Case 2
1 Real Solution
Vertex touches at one point (\(x \neq 0\)).
10
One \( x \)-intercept (at \( x = 10 \))
Case 3
0 Real Solutions
Curve never touches \( x \)-axis.
No \( x \)-intercepts
Quick Check: State whether each situation has 0, 1, or 2 Real Solutions:
1. A parabola with vertex at \( (3, 0) \) 1 real sol.
2. A parabola opening upward with vertex at \( (0, 5) \) 0 real sol.
3. A parabola that crosses at \( x = -3 \) and \( x = 7 \) 2 real sol.
4. A table where \( y \) is always negative numbers 0 real sol.
Heads Up: What About Non-Real Solutions? When a graph has 0 real solutions, does that mean the equation has NO solutions at all? In Algebra 2, you will discover that equations like \( x^2 = -4 \) actually have two complex / imaginary solutions involving \( i = \sqrt{-1} \). We are not learning how to calculate those yet—so for right now, on our real graph, we state that there are 0 Real Solutions!
Part 5
Word Problems: Deciding Between Roots & Vertex
Translator Guide: What is the question asking you to find?
Finding the ROOTS (\( y = 0 \)):
Asked when an object hits the ground, lands, enters/exits water, or when a business breaks even (\(\text{profit} = 0\)).
Finding the VERTEX (\( h, k \)):
Asked for maximum height, highest peak, turnaround point, or maximum profit. Located halfway between the roots!
Scenario 1
The Oceanarium Acrobatic Dolphin Leap
Projectile Modeling
During an oceanarium show, a marine biologist models the leap of a dolphin using the quadratic equation: h(x) = -x² + 8x - 7, where \( h(x) \) is the dolphin's height in feet above the water surface and \( x \) is the horizontal distance traveled in feet.
Step A • Set Up Standard Form for Water Level
The water surface is at height \( h(x) = 0 \). Write the equation in standard form set to zero:
-x² + 8x - 7 = 0 (or x² - 8x + 7 = 0)
Step B • Find the Roots from the Flight Table
Use the provided height table below to identify the two roots where \( h(x) = 0 \):
| Distance x (ft) |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
| Height h (ft) |
-7 |
0 |
5 |
8 |
9 |
8 |
5 |
0 |
-7 |
• Dolphin emerges from water at: x = 1 ft
• Dolphin re-enters water at: x = 7 ft
Step C • Determine the Vertex (Peak of the Leap)
1. Halfway between the roots (\( x = 1 \) and \( x = 7 \)), what is the horizontal distance? x = 4 ft (calculated as (1+7)/2 = 4)
2. What is the dolphin's maximum height at this vertex point? Height = 9 ft
3. Write the vertex as an ordered pair \( (x, y) \): (4, 9)
Step D • Solve the Trainer's Challenge Question
The trainer suspends a target ring 10 feet above the water surface directly above the middle of the leap. Does the dolphin reach the target ring? Explain using your calculations from Step C:
No, the dolphin will not reach the target ring. According to the vertex calculated in Step C, the dolphin reaches a maximum peak height of 9 feet above the water. Since 9 ft < 10 ft, the dolphin will fall 1 foot short of the ring.
Scenario 2
Robotics Club Popcorn Fundraiser
Revenue & Profit
The high school robotics team sells gourmet popcorn tins to fund their competition travel. Their business advisor models their weekly profit using the quadratic equation: P(x) = -2x² + 28x - 80, where \( x \) represents the selling price per tin (in dollars) and \( P(x) \) represents the weekly profit (in hundreds of dollars).
Step A • Break-Even Equation Setup
"Breaking even" means the club makes $0 in profit (\( P(x) = 0 \)). Write the quadratic equation that must be solved to find the break-even selling prices:
-2x² + 28x - 80 = 0 (or 2x² - 28x + 80 = 0)
Step B • Identify the Break-Even Roots
Examine the profit table computed for various selling prices:
| Selling Price x ($) |
$3 |
$4 |
$5 |
$6 |
$7 |
$8 |
$9 |
$10 |
$11 |
| Profit P ($100s) |
-14 |
0 |
10 |
16 |
18 |
16 |
10 |
0 |
-14 |
• Lower break-even price (\(P=0\)): $4
• Upper break-even price (\(P=0\)): $10
Step C • Determine the Maximum Profit (Vertex)
1. Halfway between the break-even prices ($4 and $10), what selling price maximizes profit? Price = $7 (halfway: (4+10)/2 = 7)
2. What is the maximum profit generated at this optimal price? (Remember \( P \) is in hundreds of dollars): $1,800 (18 × $100)
3. State the vertex coordinates: (7, 18)
Step D • Business Evaluation Question
A student suggests pricing the popcorn tins at $15 each to make more money. Based on your knowledge of quadratic graphs and the roots, explain why pricing at $15 would actually cause the club to lose money:
Pricing at $15 would cause the club to lose money because the parabola opens downward and the upper break-even root is $10. For any selling price above $10, profit drops below zero. At $15, profit is P(15) = -2(15)² + 28(15) - 80 = -450 + 420 - 80 = -110 hundred dollars (-$11,000 net loss). Customers will buy too few tins at such an inflated price!
