Possible roots are formed using the last term (12) of the given function and the 1st term (1) as your basis:
Since factors of 12 include plus or minus 1, 2, 3, 4, 6, 12,
possible rational roots include plus or minus 1/1, 2/1, 3/1, 4/1, 6,/1, 12/1 (and so on.) I will take a chance and try the possible root 6/1, or just plain 6.
Use synthetic division, with 6 as the divisor and all of the coefficients of <span>x^5+3x^4-5x^3-15x^2+4x+12 as dividend: ______________________ 6 / 1 3 -5 -15 4 12 6 54 -------------------------------- 1 9 49 This is not going to work; 6 is not a root.
The function y = sec(x) shifted 3 units left and 7 units down .
Step-by-step explanation:
Given the function: y = sec(x)
If k is any positive real number, then the graph of f(x) - k is the graph of y = f(x) shifted downward k units.
If p is a positive real number, then the graph of f(x+p) is the graph of y=f(x) shifted to the left punits.
The function comes from the base function y= sec(x).
Since 3 is added added on the inside, this is a horizontal shift Left 3 unit, and since 7 is subtracted on the outside, this is a vertical shift down 7 units.
Therefore, the transformation on the given function is shifted 3 units left and 7 units down