Introduction As is true with linear and exponential functions, we can perform operations on quadratic functions. Such operations include addition, subtraction, multiplication, and division. This lesson will focus on adding, subtracting, multiplying, and dividing functions to create new functions. The lesson will also explore the effects of dividing a quadratic by one of its linear factors : Operating on Functions

Key Concepts Operations with Functions Functions can be added, subtracted, multiplied, and divided. For two functions f(x) and g(x), the addition of the functions is represented as follows: For two functions f(x) and g(x), the subtraction of the functions is represented as follows: : Operating on Functions

Key Concepts, continued For two functions f(x) and g(x), the multiplication of the functions is represented as follows: For two functions f(x) and g(x), the division of the functions is represented as follows: : Operating on Functions

Key Concepts, continued Adding and subtracting linear expressions from a quadratic will yield a quadratic. Multiplying and dividing a quadratic by anything other than a constant will not yield a quadratic : Operating on Functions

Key Concepts, continued Restricted Domains When considering the division of a quadratic by a linear factor, it is possible to create a linear expression with a restricted domain. For example: For f(x) = x 2 + 5x + 6 and g(x) = x + 3, can be found such that : Operating on Functions

Key Concepts, continued In simpler terms, Remember that the denominator of a fraction cannot equal 0. Set the denominator equal to 0 and solve for x to find the restricted value(s) in the domain: x + 3 = 0, so x ≠ – : Operating on Functions

Key Concepts, continued Given the similar function h(x) = x + 2, the domain is all real numbers, and the range is the same. However, since f(x) is divided by g(x), the domain of from the preceding example is all real numbers except for x = –3 and the range is all real numbers except for y = – : Operating on Functions

Key Concepts, continued This is because when the restricted value of the domain (–3) is substituted into the simplified form of and solved for y, we get: Therefore, since x ≠ –3, then y ≠ – : Operating on Functions

Common Errors/Misconceptions forgetting to restrict the domain when dividing functions not realizing that functions must be of the same variable for like terms to be combined having difficulty moving from the formal notation to a workable problem where functions can be used with operations : Operating on Functions

Guided Practice Example 1 Let f(x) = x 2 – 3x + 4 and g(x) = x 2 + 6x – 3. Build a new function, h(x), for which h(x) = (f + g)(x) : Operating on Functions

Guided Practice: Example 1, continued 1.Expand the new function, h(x), into a form where substitution can be used. h(x) = (f + g)(x) = f(x) + g(x) The new function is expanded as h(x) = f(x) + g(x) : Operating on Functions

Guided Practice: Example 1, continued 2.Add the functions. The new function is h(x) = 2x 2 + 3x : Operating on Functions f(x) = x 2 – 3x + 4 and g(x) = x 2 + 6x – 3 Given functions from problem statement h(x) = f(x) + g(x)Expanded notation h(x) = (x 2 – 3x + 4) + (x 2 + 6x – 3) Substitute values for f(x) and g(x). h(x) = 2x 2 + 3x + 1Combine like terms. ✔

Guided Practice: Example 1, continued : Operating on Functions

Guided Practice Example 3 For f(x) = 3x x – 10 and g(x) = x + 5, find What type of function is the quotient of Are there restrictions on the domain and range of the function : Operating on Functions

Guided Practice: Example 3, continued 1.Since the functions are being divided, write the functions f(x) and g(x) as a fraction : Operating on Functions

Guided Practice: Example 3, continued 2.Factor the quadratic function, f(x) : Operating on Functions

Guided Practice: Example 3, continued 3.Simplify the equation and define the type of equation of the simplified form. Divide away the monomial (x + 5) from the top and bottom of the fraction: The function is a linear equation that graph the line y = 3x – : Operating on Functions

Guided Practice: Example 3, continued 4.Look at the original fraction to see if there are restricted values on the domain. In this case, x ≠ –5 because (–5) + 5 = 0 and division by 0 is undefined. Next, substitute x = –5 into the final equation to determine the restricted value(s) of y. 3x – 2 = 3(–5) – 2 = –17 Since x ≠ –5, then y ≠ – : Operating on Functions ✔

Guided Practice: Example 3, continued : Operating on Functions