\((x^2 +5x-6 )\over(x^2+x-6)\)
From this factoring, it can be deduced that the function has vertical asymptotes (when the denominator equals 0) at x=3 and x=-2. Also, it can be determined that the function has x-intercepts (when the numerator is zero and the denominator isn't) at x=-6 and x=1.
From this data, it can be concluded that the correct answer is the second graph.
P.s. You can never cross a vertical asymptote.
Think of this problem as an interest problem. After a certain amount of time, your money (cells) will double.
The standard equation for simple compounding interest is y = P(R)^x
P is the principal/ initial amount
R is the rate at which your principle will increase
x is the time interval (seconds, days, year, etc.)
y is the final amount after x time
Knowing that we can make an equation
where h is (t/14)
t is in hours so every fourteen hours the expression h increases by 1 and so the equation N doubles.
There are 24 hours in a day so in 2 days that is 48 hours
h will then be 48/14
plug this into the equation and you get
N= 1000(2)^(48/14) = 10767 cells (rounding down because you can't have a portion of a cell)
The equation for the perimeter of a square is P=4d where d is the length of one side
To find the distance between two points on a cartesian plane you use the distance formula
\(d = \sqrt((x2-x1)^2+(y2-y1)^2)\)
\(d = \sqrt((5-1)^2+(-1-4)^2)\)Where B is (x2,y2) and A is (x1,y1)
sqrt(41) is about 6.4
so we plug that into the perimeter equation and get P=25.6
There is a zero at 2 with multiplicity 3
\(f(x) = (x-2)^3 (?)\)
A zero at 6 with multiplicity 1
A zero of 5i
It is good to note that complex factors come in pairs so by adding a complex factor you essentially solve for two complex zeroes
Now we have a polynomial with a degree of 6. However, we need the leading coefficient to be 10. To do this we just include a ten as a subset of the function like this. \(f(x)=10(x-2)^3(x-6)(x^2+25)\) Now you have a polynomial f(x) that satisfies all the given conditions.