This actually should be reported. There are no choices. However there are ways of finding out what the zeros are and I will do that first to begin with.
The first is a bit of a cheat. My calculator will do it, so I'm just going to get the answer from it.
x can equal 3
x can be 5
x can be 4
So f(x)= a (x - 3)(x-5 )(x -4) is the basic equation. You need another point other than the zeros to determine the value of a.
Method 2
Get the graph.
I've done that below. Basically it agrees with what my calculator says.
Method 3
You could factor it, but simple factoring methods don't work with this question.
Method 4
You could do this in a method similar to a quadratic, but I can't do that. It is on my to do list to lean though.
The first is a bit of a cheat. My calculator will do it, so I'm just going to get the answer from it.
x can equal 3
x can be 5
x can be 4
So f(x)= a (x - 3)(x-5 )(x -4) is the basic equation. You need another point other than the zeros to determine the value of a.
Method 2
Get the graph.
I've done that below. Basically it agrees with what my calculator says.
Method 3
You could factor it, but simple factoring methods don't work with this question.
Method 4
You could do this in a method similar to a quadratic, but I can't do that. It is on my to do list to lean though.
