Algebra 2 Standard Form of a Quadratic Function

Algebra 2 Standard Form of a Quadratic Function Lesson 4 -2 Part 2

Goals Goal • To graph quadratic functions written in standard form. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems.

Vocabulary • None

Essential Question Big Idea: Function and Equivalence • When is vertex form more useful than standard form?

Quadratic Function - Forms Sometimes it is useful to change the form of the quadratic function. Compare the two forms. Standard Form Vertex Form • f(x) = ax 2 + bx + c • f(x) = a(x – h)2 + k • Vertex: (h, k) • Axis of Symmetry: • y-intercept: (0, c) • Axis of Symmetry: x = h a > 0 parabola opens up a < 0 parabola opens down a is the same in both forms

Convert from Vertex Form to Standard Form f(x) = a(x – h)2 + k f(x) = ax 2 + bx + c 1. Square the binomial (x – h)2. 2. Distribute a. 3. Combine like terms. 4. Simplify.

Example: • Convert f(x) = 3(x – 1)2 + 8 to standard form. = 3(x – 1) + 8 = 3(x 2 – x + 1) + 8 = 3(x 2 – 2 x + 1) + 8 = 3 x 2 – 6 x + 3 + 8 f(x) = 3 x 2 – 6 x + 11

Your Turn: 1. Convert f(x) = 3(x + 1)2 – 5 to standard form. Standard form: f(x) = 3 x 2 + 6 x – 2 2. Convert y = 2(x – 4)2 – 1 to standard form. Standard form: y = 2 x 2 – 16 x – 31

Convert from Standard Form to Vertex Form f(x) = ax 2 + bx + c f(x) = a(x – h)2 + k 1. Calculate the x-coordinate of the vertex (h): 2. Substitute the x-coordinate of the vertex into the function to calculate the y-coordinate of the vertex (k): 3. Use a from the standard form. 4. Substitute a, h, and k into vertex form.

Example: Convert f(x) = -3 x 2 – 12 x – 13 to vertex form. 1. (h = -2) 2. y = -3(-2)2 – 12(-2) – 13 y = -12 + 24 -13 y = -1 3. From standard form a = -3. 4. Substitute f(x) = a(x – h)2 + k (k = -1) f(x) = -3(x + 2)2 – 1

Example: Convert f(x) = -3 x 2 + 9 x – 1 to vertex form. 1. 2. 3. 4. From standard form a = -3. Substitute f(x) = a(x – h)2 + k

Your Turn: Convert y = 2 x 2 + 8 x + 4 to vertex form. 1. (h = -2) 2. y = 2(-2)2 + 8(-2) + 4 y = 8 – 16 + 4 y = -4 3. From standard form a = 2. 4. Substitute y = a(x – h)2 + k (k = -4) y = 2(x + 2)2 – 4

Your Turn: • Convert f(x) = 2 x 2 – 4 x + 5 to vertex form. Vertex = (1, 3) a=2 Vertex form: f(x) = 2(x-1)2 + 3

Quadratic Function Applications • When a soccer ball is kicked into the air, how long will the ball take to hit the ground? • The height h in feet of the ball after t seconds can be modeled by the quadratic function h(t) = – 16 t 2 + 32 t. • In this situation, the value of the function represents the height of the soccer ball. And the y-coordinate, h(t), of the vertex represents the maximum height of the ball. • Quadratic application problems often involve finding the minimum or maximum of the quadratic function.

Important! The minimum (or maximum) value is the y-value at the vertex. It is not the ordered pair that represents the vertex.

Example: Find the minimum or maximum value of f(x) = – 3 x 2 + 2 x – 4. Then state the domain and range of the function. Step 1 Determine whether the function has minimum or maximum value. Because a is negative, the graph opens downward and has a maximum value. Step 2 Find the x-value of the vertex. Substitute 2 for b and – 3 for a.

Continued f(x) = – 3 x 2 + 2 x – 4 Step 3 Then find the y-value of the vertex, The maximum value is. The domain is all real numbers, R. The range is all real numbers less than or equal to

Your Turn: Find the minimum or maximum value of f(x) = x 2 – 6 x + 3. Then state the domain and range of the function. Step 1 Determine whether the function has minimum or maximum value. Because a is positive, the graph opens upward and has a minimum value. Step 2 Find the x-value of the vertex.

Continued f(x) = x 2 – 6 x + 3 Step 3 Then find the y-value of the vertex, f(3) = (3)2 – 6(3) + 3 = – 6 The minimum value is – 6. The domain is all real numbers, R. The range is all real numbers greater than or equal to – 6, or {y|y ≥ – 6}.

Example: Application The average height h in centimeters of a certain type of grain can be modeled by the function h(r) = 0. 024 r 2 – 1. 28 r + 33. 6, where r is the distance in centimeters between the rows in which the grain is planted. Based on this model, what is the minimum average height of the grain, and what is the row spacing that results in this height?

Continued The average height h in centimeters of a certain type of grain can be modeled by the function h(r) = 0. 024 r 2 – 1. 28 r + 33. 6, where r is the distance in centimeters between the rows in which the grain is planted. Based on this model, what is the minimum average height of the grain, and what is the row spacing that results in this height? The minimum value will be at the vertex (r, h(r)). Step 1 Find the r-value of the vertex using a = 0. 024 and b = – 1. 28.

Continued Step 2 Substitute this r-value into h to find the corresponding minimum, h(r) = 0. 024 r 2 – 1. 28 r + 33. 6 Substitute 26. 67 for r. h(26. 67) = 0. 024(26. 67)2 – 1. 28(26. 67) + 33. 6 h(26. 67) ≈ 16. 5 The minimum height of the grain is about 16. 5 cm planted at 26. 7 cm apart.

Your Turn: The highway mileage m in miles per gallon for a compact car is approximately by m(s) = – 0. 025 s 2 + 2. 45 s – 30, where s is the speed in miles per hour. What is the maximum mileage for this compact car to the nearest tenth of a mile per gallon? What speed results in this mileage?

Solution The highway mileage m in miles per gallon for a compact car is approximately by m(s) = – 0. 025 s 2 + 2. 45 s – 30, where s is the speed in miles per hour. What is the maximum mileage for this compact car to the nearest tenth of a mile per gallon? What speed results in this mileage? The maximum value will be at the vertex (s, m(s)). Step 1 Find the s-value of the vertex using a = – 0. 025 and b = 2. 45 ) ( b s === 49 2 a 2 (-0. 025)

Solution Step 2 Substitute this s-value into m to find the corresponding maximum, m(s) = – 0. 025 s 2 + 2. 45 s – 30 Substitute 49 for r. m(49) = – 0. 025(49)2 + 2. 45(49) – 30 m(49) ≈ 30 The maximum mileage is 30 mi/gal at 49 mi/h.

Assignment • Section 4 -2 part 2, Pg 218 #5 -9