we have
Convert into vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
Complete the square. Remember to balance the equation by adding the same constants to each side
Rewrite as perfect squares
-------> equation in vertex form
This is the equation of a vertical parabola open down
The vertex is a maximum--------->
therefore
the domain of the function is the interval---------> (-∞,∞)
Domain is all real numbers
The range of the function is the interval --------> (-∞,4]
Range is all real number less than or equal to
Using a graphing tool
see the attached figure to better understand the problem
one way would be to factor
I can't factor it so we will have to use Descartes' Rule of Signs which is helpful for finding how many real roots you have
it goes like this:
for a polynomial with real coefients, consider .
after arranging the terms in decending order in terms of degree, count how many times the signs of the coeffients change direction and minus 2 from that number until you get to 1 or 0. that will be the number of even roots the function can have
We have (-, -, +, +). the signs changed 1 times, so it has 1 real positive root
to get the negative roots, we evaluate f(-x) and see how many times the root changes
signs are (-, -, -, +). there was 1 change in sign
so the function has 1 real negative root
a total of 2 real roots
a function of degree can have at most,
roots
our function is degree 6 so it has 6 roots
if 2 are real, then the others must be complex
6-2=4 so there are 4 complex roots
you can also show that there are only 2 real roots by using a graphing utility to see that there are only 2 real roots
