Evaluating Functions Assignment Algebra 1

David W.

Round Rock, TX

I️ don’t know how to do this

H(a)=a2+5a; find h(8)

Feb 5 | Tiffany from Winslow, AR | 1 Answer | 0 Votes

Evaluating Functions

Kenneth S.

Mesa, AZ

how would be find the answer to f(g(x)) when f(x)=2x-12 , g(x)=x^2+2x+4, f(x)=-3/2x-6

f(x)=2x-12 , g(x)=x2+2x+4, f(x)=-3/2x-6

Oct 7 | Katie from Centreville, VA | 1 Answer | 0 Votes

Evaluating Functions

Michael J.

East Elmhurst, NY

Find the initial amount injected and the amount in the bloodstream after 7 hours. Round your answers to the nearest hundredth as necessary

The number of milligrams D(h) of a drug in a patient's bloodstream h hours after the drug is injected is modeled by the following function. D(h)=25e^-0.3h

Apr 15 | Asterly from Pompano Beach, FL | 1 Answer | 0 Votes

Evaluating Functions

Kenneth S.

Mesa, AZ

Evaluate the problem

If A?B=A^2+AB+B^2, evaluate 4?3

8/24/2016 | Lillianah from Cleveland, OH | 2 Answers | 0 Votes

Evaluating Functions

Michael J.

East Elmhurst, NY

Find and simplify: f(-x) - f(x) for the given function of f

f(x)= 5x^3 + 6x^2 - 3x - 17   I have tried this problem many times and cannot seem to get the correct answer.   Please help!   Thanks!

10/20/2015 | Erin from East Lansing, MI | 1 Answer | 0 Votes

FunctionsEvaluating Functions

Michael F.

Wilton, CT

Find and simplify the difference quotient

Find and simplify the difference quotient: f(x+h) - f(x) / h, where h cannot = 0, for the function: f(x)= 3x2 - x +2   Have tried this multiple times and still get the same answer...

10/20/2015 | Erin from East Lansing, MI | 1 Answer | 0 Votes

FunctionsEvaluating Functions

Bruce Y.

Cincinnati, OH

Evaluate the function at the given values of the independent variable and simplify.

f(x)= (3x)/(11*lxl)   (That's the absolute value of x)   f(r^8)   I don't understand how you get the answers after plugging in r^8 in the equation

FunctionsEvaluating Functions

Mark M.

Carson, CA

Evaluate enter only the sum

S21for 23+20+17+14+.....

4/7/2015 | Kyrsten from Fort Wayne, IN | 1 Answer | 0 Votes

Algebra 2HelpfulEvaluating FunctionsSequencesMath

Phillip R.

North Providence, RI

how do i evaluate this function?

k(n)=3n-4; find k(-2-n)

8/26/2014 | Rylee from Cheyenne, WY | 1 Answer | 0 Votes

Algebra 2Evaluating Functions

Philip P.

Olney, MD

how do i evaluate this function?

g(x) =3x-3 Find g(6)

8/26/2014 | Rylee from Cheyenne, WY | 1 Answer | 0 Votes

Algebra 2Evaluating Functions

Evaluating Functions

Evaluating Functions

To evaluate a function is to:

Replace (substitute) its variable with a given number or expression.

Like in this example:

Example: evaluate the function f(x) = 2x+4 for x=5

Just replace the variable "x" with "5":

f(5) = 2×5 + 4 = 14

Answer: f(5) = 14

More Examples

Here is a function:

f(x) = 1 − x + x2

Important! The "x" is just a place-holder! And "f" is just a name.

These are all the same function:

  • f(x) = 1 − x + x2
  • f(q) = 1 − q + q2
  • w(A) = 1 − A + A2
  • pumpkin(θ) = 1 − θ + θ2

 

Evaluate For a Given Value:

Let us evaluate that function for x=3:

f(3) = 1 − 3 + 32 = 1 − 3 + 9 = 7

Evaluate For a Given Expression:

Evaluating can also mean replacing with an expression (such as 3m+1 or v2).

Let us evaluate the function for x=1/r:

f(1/r) = 1 − (1/r) + (1/r)2

Or evaluate the function for x = a−4:

f(a−4) = 1 − (a−4) + (a−4)2

 = 1 − a + 4 + a2 − 8a + 16

 = 21 − 9a + a2

Another Example

You can use your ability to evaluate functions to find other answers:

Example: h(x) = 3x2 + ax − 1

  • You are told that h(3) = 8, can you work out what "a" is?

 

First, evaluate h(3):h(3) = 3×(3)2 + a×3 − 1

Simplify:h(3) = 27 + 3a − 1

 h(3) = 26 + 3a

 

Now ... we know that h(3) = 8, so: 8 = 26 + 3a

Swap sides:26 + 3a = 8

Subtract 26 from both sides:3a = −18

Divide by 3:a = −6

 

Check: h(3) = 3(3)2 − 6×3 − 1 = 27 − 18 − 1 = 8

Careful!

I recommend putting the substituted values inside parentheses () , so you don't make mistakes.

Example: evaluate the function h(x) = x2 + 2 for x = −3

Replace the variable "x" with "−3":

h(−3) = (−3)2 + 2 = 9 + 2 = 11

Without the () you could make a mistake:

h(−3) = −32 + 2 = −9 + 2 = −7 (WRONG!)

Also be careful of this:

f(x+a) is not the same as f(x) + f(a)

Example: g(x) = x2

g(w+1) = (w+1)2 = w2 + 2w + 1

vs

g(w) + g(1) = w2 + 12 = w2 + 1

Different Result!

 

 

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