All Algebra II Resources
Example Questions
Example Question #11 : Function Notation
Given , find .
Plug in a for x:
Next plug in (a + h) for x:
Therefore f(a+h) - f(a) = .
Example Question #82 : Functions And Lines
Solve for when .
Does not exist
Plug 3 in for x:
Simplify:
=
= 5
Example Question #3374 : Algebra 1
What is of the following equation?
To complete an equation with a function, plug the number inside the parentheses into the equation for and solve algebraically.
In this case the
Square the 7 and multiply to get
Add the numbers to get the answer .
Example Question #3375 : Algebra 1
Example Question #3376 : Algebra 1
Example Question #13 : Function Notation
If , find .
If then can be rewritten as . Therefore, . Subtracting from both sides of the eqauation gives .
Example Question #661 : Algebra Ii
Evaluate the function for .
To solve for the value of the function at , simply plug in the value in place of every . By doing this, you will be left with the equation:.
Another way to go about the problem is first simplifying the expression so that like terms are collected, so . Then to find , simply plug in the value in place of every . By doing this, you will be left with the equation:.
Example Question #671 : Algebra Ii
Given and , find .
Undefined
Given f(x) and g(x), find f(g(5))
This type of problem can look intimidating depending on how it is set up. What it is asking is for us to plug g(x) into f(x) everywhere we see an x, and then to plug in 5 everywhere we still have an x. It gets a little cumbersome if approached all at once:
This looks a bit unwieldy, but this problem can be approached easily by looking at it in layers.
First, find g(5)
.
Next, plug that 15 into f(x).
So our answer is:
Example Question #672 : Algebra Ii
If and , what is ?
We start by replacing all the 's in with the entire function of :
We then FOIL the first term:
And we collect all the like terms (the constants in this case):
Example Question #17 : Function Notation
What is
?
To find the composition of two functions, substitute the second equation in to the first function.
Therefore,
and
Thus,
.