To put an equation into (x+c)^2, we need to see if the trinomial is a perfect square.
General form of a trinomial: ax^2+bx+c
If c is a perfect square, for example (1)^2=1, 2^2=4, that's a good indicator that it's a perfect square trinomial.
Here, it is, because 1 is a perfect square.
To ensure that it's a perfect square trinomial, let's look at b, which in this case is 2.
It has to be double what c is.
2 is the double of 1, therefore this is a perfect square trinomial.
Knowing this, we can easily put it into the form (x+c)^2.
And the answer is: (x+1)^2.
To do it the long way:
x^2+2x+1
Find 2 numbers that add to 2 and multiply to 1.
They are both 1.
x^2+x+x+1
x(x+1)+1(x+1)
Gather like terms
(x+1)(x+1)
or (x+1)^2.
Option 3, is correct.
Just find the one that have the x's only, not the one that is squared or without one. It doesn't matter if the term is a negative or positive.
If you have any further questions, please reach out to me :)
Part a)
a) The given function is
We let
Interchange x and y.
Solve for y;
Part b) The range of f(x) refers to y-values for which f(x) exists.
The range of f(x) is
This is because the function is within y=-3 and y=3.
c) The range of
is
The domain is -3≤x≤3
This is because the domain and range of a function and its inverse swaps.
Part d) The graph is shown in the attachment.
If you times 39.99 by 0.20 (20%) then you get your answer of $7.99
