If quadrilateral JKLM had a translation where all the vertices were in the second quadrant, what would be the vertices?
2 answers:
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To me personally, the first bit f(g(x)) is easy and the domain is tricky. Let's try explain this.
A function takes an input number and returns an output number depending on the function. Look at f(x) = x+3, if we let the input number be 2 then we say that f(2) = 5. We could do f(π) to give us π+3 or even f(x²) to give us x² +3. The trick is to substitute the input into the function equation.
You have been asked to find f(g(x)). You know f(x) =
. Putting numbers in at this point would be easy (try work out f(2), you'll do it really quick) but you have to put in g(x).
f(g(x)) =
we also know that g(x) =
so we can say that
f(g(x)) =
and that is f(g(x)) but the question requires that we simplify it so
f(g(x)) =
Now for the tricky bit (for me, at least). The domain is the full set of values that you can 'put in to' the function and still get a real value out. So how do we work out what numbers 'break' the function? I like to use the fact that DIVIDING BY ZERO IS IMPOSSIBLE. What value of x can we put into the function to make it so the function is being divided by 0? i.e. 1-2x = 0 solve that and you have a value of x that isn't part of the domain.
This means the domain is all real numbers EXCEPT the solution to that equation. (Because if we put that value into f(g(x)) it's impossible to get a value out.)
[I know this was a lot to read, if you have any questions or don't get anything feel free to message me or leave a comment.]
A function takes an input number and returns an output number depending on the function. Look at f(x) = x+3, if we let the input number be 2 then we say that f(2) = 5. We could do f(π) to give us π+3 or even f(x²) to give us x² +3. The trick is to substitute the input into the function equation.
You have been asked to find f(g(x)). You know f(x) =
f(g(x)) =
we also know that g(x) =
f(g(x)) =
f(g(x)) =
Now for the tricky bit (for me, at least). The domain is the full set of values that you can 'put in to' the function and still get a real value out. So how do we work out what numbers 'break' the function? I like to use the fact that DIVIDING BY ZERO IS IMPOSSIBLE. What value of x can we put into the function to make it so the function is being divided by 0? i.e. 1-2x = 0 solve that and you have a value of x that isn't part of the domain.
This means the domain is all real numbers EXCEPT the solution to that equation. (Because if we put that value into f(g(x)) it's impossible to get a value out.)
[I know this was a lot to read, if you have any questions or don't get anything feel free to message me or leave a comment.]
