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Algebra 1

6.1.3 Evaluating a Polynomial Function for a Given Value

Algebra 16.1.3 Evaluating a Polynomial Function for a Given Value

6.1.3 Evaluating a Polynomial Function for a Given Value

6.1.3 • Evaluating a Polynomial Function for a Given Value

Activity

Evaluate each function for h ( − 4 ) h ( − 4 ) , h ( 0 ) h ( 0 ) , and h ( 6 ) h ( 6 ) .

1. h ( x ) = 3 x 2 − 7 x − 9 h ( x ) = 3 x 2 − 7 x − 9

2. h ( x ) = − 4 x 2 − 3 x + 2 h ( x ) = − 4 x 2 − 3 x + 2

3. h ( x ) = − 2 x 2 − 5 x + 15 h ( x ) = − 2 x 2 − 5 x + 15

Evaluate each function for d ( − 2 ) d ( − 2 ) , d ( 0 ) d ( 0 ) , and d ( 3 ) d ( 3 ) .

4. d ( x ) = 8 x 2 − 2 x + 6 d ( x ) = 8 x 2 − 2 x + 6

5. d ( x ) = − 5 x 2 − 4 x + 10 d ( x ) = − 5 x 2 − 4 x + 10

6. d ( x ) = − 3 x 2 + 6 x + 7 d ( x ) = − 3 x 2 + 6 x + 7

7. Work with a partner. One partner should write their own polynomial function similar to the ones in this activity. The other partner should select 3 different values between –10 and 10. Exchange your function and values with your partner. Each member should evaluate the polynomial function for the given values. When finished, compare your answers. Did you find the same solutions? Share any differences you found.

Self Check

For the function g ( x ) = 5 x 2 − 8 x + 4 , find g ( − 3 ) .
  1. 73
  2. 25
  3. 13
  4. − 17

Additional Resources

Evaluating a Polynomial Function for a Given Value

A polynomial function is a function whose range values are defined by a polynomial. For example, f ( x ) = x 2 + 5 x + 6 f ( x ) = x 2 + 5 x + 6 and g ( x ) = 3 x − 4 g ( x ) = 3 x − 4 are polynomial functions because x 2 + 5 x + 6 x 2 + 5 x + 6 and 3 x − 4 3 x − 4 are polynomials.

To evaluate a polynomial function, we will substitute the given value for the variable and then simplify using the order of operations.

Let’s look at some examples.

For the function f ( x ) = 5 x 2 − 8 x + 4 f ( x ) = 5 x 2 − 8 x + 4 , find:

  1. f ( 4 ) f ( 4 )
  2. f ( − 2 ) f ( − 2 )
  3. f ( 0 ) f ( 0 )

The following is a step-by-step breakdown.

1. f ( x ) = 5 x 2 − 8 x + 4 f ( x ) = 5 x 2 − 8 x + 4

Step 1 - To find f ( 4 ) f ( 4 ) , substitute 4 4 for x x .

f ( 4 ) = 5 ( 4 ) 2 − 8 ( 4 ) + 4 f ( 4 ) = 5 ( 4 ) 2 − 8 ( 4 ) + 4

Step 2 - Simplify the exponents.

f ( 4 ) = 5 · 16 − 8 ( 4 ) + 4 f ( 4 ) = 5 · 16 − 8 ( 4 ) + 4

Step 3 - Multiply.

f ( 4 ) = 80 − 32 + 4 f ( 4 ) = 80 − 32 + 4

Step 4 - Simplify.

f ( 4 ) = 52 f ( 4 ) = 52

2. f ( x ) = 5 x 2 − 8 x + 4 f ( x ) = 5 x 2 − 8 x + 4

Step 1 - To find f ( − 2 ) f ( − 2 ) , substitute − 2 − 2 for x x .

f ( − 2 ) = 5 ( − 2 ) 2 − 8 ( − 2 ) + 4 f ( − 2 ) = 5 ( − 2 ) 2 − 8 ( − 2 ) + 4

Step 2 - Simplify the exponents.

f ( − 2 ) = 5 · 4 − 8 ( − 2 ) + 4 f ( − 2 ) = 5 · 4 − 8 ( − 2 ) + 4

Step 3 - Multiply.

f ( − 2 ) = 20 + 15 + 4 f ( − 2 ) = 20 + 15 + 4

Step 4 - Simplify.

f ( − 2 ) = 40 f ( − 2 ) = 40

3. f ( x ) = 5 x 2 − 8 x + 4 f ( x ) = 5 x 2 − 8 x + 4

Step 1 - To find f ( 0 ) f ( 0 ) , substitute 0 0 for x x .

f ( 0 ) = 5 ( 0 ) 2 − 8 ( 0 ) + 4 f ( 0 ) = 5 ( 0 ) 2 − 8 ( 0 ) + 4

Step 2 - Simplify the exponents.

f ( 0 ) = 5 · 0 − 8 ( 0 ) + 4 f ( 0 ) = 5 · 0 − 8 ( 0 ) + 4

Step 3 - Multiply.

f ( 0 ) = 0 + 0 + 4 f ( 0 ) = 0 + 0 + 4

Step 4 - Simplify.

f ( 0 ) = 4 f ( 0 ) = 4

Try it

Try It: Evaluating a Polynomial Function for a Given Value

For the function f ( x ) = 3 x 2 + 2 x − 15 f ( x ) = 3 x 2 + 2 x − 15 , find:

1. f ( 3 ) f ( 3 )

2. f ( − 5 ) f ( − 5 )

3. f ( 0 ) f ( 0 )

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