2.4 Quadratic Graphs
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Learning Objectives
- Understand the general shape of a quadratic graph
- Identify key features of quadratic graphs including the vertex, axis of symmetry, and intercepts
- Sketch quadratic graphs from their equations
- Determine the equation of a quadratic graph from given information
- Solve problems involving quadratic graphs
Key Concepts
Introduction to Quadratic Graphs
A quadratic graph, or parabola, is the graph of a quadratic function f(x) = ax² + bx + c, where a, b, and c are constants and a ≠ 0.
The general shape of a quadratic graph is a U-shape (if a > 0) or an inverted U-shape (if a < 0).
Quadratic graphs are symmetric about a vertical line called the axis of symmetry, which passes through the vertex of the parabola.
Key Features of Quadratic Graphs
The vertex is the highest point (if a < 0) or lowest point (if a > 0) of the parabola. It has coordinates (h, k) when the quadratic is written in the form f(x) = a(x - h)² + k.
The axis of symmetry is the vertical line x = h, where h is the x-coordinate of the vertex.
The y-intercept is the point where the graph crosses the y-axis, which occurs at (0, c).
The x-intercepts, if they exist, are the points where the graph crosses the x-axis, which occur when f(x) = 0.
Example 1
Identify the key features of the quadratic graph f(x) = x² - 6x + 8.
Solution:
Step 1: Find the vertex by completing the square. f(x) = x² - 6x + 8 = (x - 3)² - 1 So the vertex is at (3, -1). Step 2: The axis of symmetry is the vertical line x = 3. Step 3: Find the y-intercept by evaluating f(0). f(0) = 0² - 6(0) + 8 = 8 So the y-intercept is at (0, 8). Step 4: Find the x-intercepts by solving f(x) = 0. x² - 6x + 8 = 0 (x - 2)(x - 4) = 0 x = 2 or x = 4 So the x-intercepts are at (2, 0) and (4, 0).
Example 2
Identify the key features of the quadratic graph f(x) = -2x² + 8x - 5.
Solution:
Step 1: Find the vertex by completing the square. f(x) = -2x² + 8x - 5 = -2(x² - 4x) - 5 = -2(x - 2)² + 3 So the vertex is at (2, 3). Step 2: The axis of symmetry is the vertical line x = 2. Step 3: Find the y-intercept by evaluating f(0). f(0) = -2(0)² + 8(0) - 5 = -5 So the y-intercept is at (0, -5). Step 4: Find the x-intercepts by solving f(x) = 0. -2x² + 8x - 5 = 0 Using the quadratic formula with a = -2, b = 8, c = -5: x = (-8 ± √(64 - 40))/(-4) = (-8 ± √24)/(-4) = (-8 ± 2√6)/(-4) = (8 ± 2√6)/4 = 2 ± (√6)/2 So the x-intercepts are at (2 + (√6)/2, 0) and (2 - (√6)/2, 0), which are approximately (3.22, 0) and (0.78, 0).
Sketching Quadratic Graphs
To sketch a quadratic graph, follow these steps:
1. Identify whether the parabola opens upward (a > 0) or downward (a < 0).
2. Find the vertex by completing the square or using the formula x = -b/(2a).
3. Find the y-intercept by evaluating f(0).
4. Find the x-intercepts (if they exist) by solving f(x) = 0.
5. Plot these key points and draw a smooth curve through them, ensuring the correct shape.
Example 1
Sketch the graph of f(x) = x² - 4x + 3.
Solution:
Step 1: Since a = 1 > 0, the parabola opens upward. Step 2: Find the vertex by completing the square. f(x) = x² - 4x + 3 = (x - 2)² - 1 So the vertex is at (2, -1). Step 3: Find the y-intercept. f(0) = 0² - 4(0) + 3 = 3 So the y-intercept is at (0, 3). Step 4: Find the x-intercepts by solving f(x) = 0. x² - 4x + 3 = 0 (x - 1)(x - 3) = 0 x = 1 or x = 3 So the x-intercepts are at (1, 0) and (3, 0). Step 5: Plot these points and draw a smooth U-shaped curve through them, with the vertex at the lowest point and the curve symmetric about the line x = 2.
Example 2
Sketch the graph of f(x) = -x² + 2x + 3.
Solution:
Step 1: Since a = -1 < 0, the parabola opens downward. Step 2: Find the vertex by completing the square. f(x) = -x² + 2x + 3 = -(x² - 2x) + 3 = -(x - 1)² + 4 So the vertex is at (1, 4). Step 3: Find the y-intercept. f(0) = -(0)² + 2(0) + 3 = 3 So the y-intercept is at (0, 3). Step 4: Find the x-intercepts by solving f(x) = 0. -x² + 2x + 3 = 0 x² - 2x - 3 = 0 (x - 3)(x + 1) = 0 x = 3 or x = -1 So the x-intercepts are at (3, 0) and (-1, 0). Step 5: Plot these points and draw a smooth inverted U-shaped curve through them, with the vertex at the highest point and the curve symmetric about the line x = 1.
Finding the Equation of a Quadratic Graph
To find the equation of a quadratic graph, you need sufficient information about the graph, such as:
- The coordinates of the vertex and one other point
- Three points that lie on the graph
- The coordinates of the vertex and the direction of opening (up or down)
Example 1
Find the equation of a quadratic graph with vertex at (2, -3) and passing through the point (4, 5).
Solution:
Step 1: Since the vertex is at (2, -3), the equation has the form f(x) = a(x - 2)² - 3, where a is a constant to be determined. Step 2: Use the given point (4, 5) to find a. 5 = a(4 - 2)² - 3 5 = a(4) - 3 8 = 4a a = 2 Step 3: Write the equation in the form f(x) = a(x - h)² + k. f(x) = 2(x - 2)² - 3 Step 4: Expand to get the standard form. f(x) = 2(x² - 4x + 4) - 3 = 2x² - 8x + 8 - 3 = 2x² - 8x + 5
Example 2
Find the equation of a quadratic graph passing through the points (1, 4), (2, 7), and (3, 12).
Solution:
Step 1: Let the equation be f(x) = ax² + bx + c, where a, b, and c are constants to be determined. Step 2: Substitute the three points into the equation to get a system of three equations. For (1, 4): a(1)² + b(1) + c = 4 → a + b + c = 4 For (2, 7): a(2)² + b(2) + c = 7 → 4a + 2b + c = 7 For (3, 12): a(3)² + b(3) + c = 12 → 9a + 3b + c = 12 Step 3: Solve the system of equations. From the first equation: c = 4 - a - b Substituting into the second equation: 4a + 2b + (4 - a - b) = 7 3a + b = 3 Substituting into the third equation: 9a + 3b + (4 - a - b) = 12 8a + 2b = 8 4a + b = 4 Solving the two equations 3a + b = 3 and 4a + b = 4: b = 4 - 4a 3a + (4 - 4a) = 3 3a - 4a + 4 = 3 -a = -1 a = 1 Then b = 4 - 4(1) = 0 and c = 4 - 1 - 0 = 3 Step 4: Write the equation in standard form. f(x) = 1x² + 0x + 3 = x² + 3
Applications of Quadratic Graphs
Quadratic graphs have many applications in various fields, including:
- Physics: modeling projectile motion, where the height of an object varies quadratically with time
- Economics: modeling cost, revenue, and profit functions
- Engineering: designing arches, bridges, and other structures
- Optimization problems: finding maximum or minimum values
Example 1
A ball is thrown vertically upward from a height of 1.5 meters with an initial velocity of 20 m/s. The height h (in meters) of the ball after t seconds is given by h = -4.9t² + 20t + 1.5. Find (a) the maximum height reached by the ball, and (b) when the ball hits the ground.
Solution:
Step 1: Find the maximum height by finding the vertex of the quadratic function. The coefficient of t² is a = -4.9 < 0, so the parabola opens downward and the vertex gives the maximum height. Using the formula t = -b/(2a) for the t-coordinate of the vertex: t = -20/(2(-4.9)) = 20/9.8 ≈ 2.04 seconds The maximum height is h(2.04) = -4.9(2.04)² + 20(2.04) + 1.5 ≈ -4.9(4.16) + 40.8 + 1.5 ≈ -20.38 + 40.8 + 1.5 ≈ 21.92 meters Step 2: Find when the ball hits the ground by solving h = 0. -4.9t² + 20t + 1.5 = 0 Using the quadratic formula with a = -4.9, b = 20, c = 1.5: t = (-20 ± √(400 - 4(-4.9)(1.5)))/(-9.8) = (-20 ± √(400 + 29.4))/(-9.8) = (-20 ± √429.4)/(-9.8) Since we're looking for when the ball hits the ground after being thrown, we need the positive value of t that is greater than 0: t ≈ (-20 + 20.72)/(-9.8) ≈ -0.07 seconds (invalid) or t ≈ (-20 - 20.72)/(-9.8) ≈ 4.16 seconds So the ball hits the ground after approximately 4.16 seconds.
Example 2
A company finds that the profit P (in thousands of dollars) from selling x hundred units of a product is given by P = -2x² + 24x - 50. Find (a) the number of units that should be sold to maximize profit, and (b) the maximum profit.
Solution:
Step 1: Find the number of units that maximizes profit by finding the vertex of the quadratic function. The coefficient of x² is a = -2 < 0, so the parabola opens downward and the vertex gives the maximum profit. Using the formula x = -b/(2a) for the x-coordinate of the vertex: x = -24/(2(-2)) = 24/4 = 6 So the company should sell 6 hundred units, or 600 units, to maximize profit. Step 2: Find the maximum profit by evaluating P(6). P(6) = -2(6)² + 24(6) - 50 = -2(36) + 144 - 50 = -72 + 144 - 50 = 22 So the maximum profit is 22 thousand dollars, or $22,000.
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