Question

1. Give the domain and range of this function:y=1(x+20)^2
2.Give the asymptotes and intercepts for this function: y=1(x-2)+2
3. identify the shifts: x=1/(x+4) -4
4. List all intervals (increasing, decreasing, positive, negative):y=(x^2)(x^2-2)-1/2
Any help is appreciated !!! thxs in Advance

Answer 1

\(\large\color{midnightblue}{ y=(x+20)^2 }\)
there is no restriction for the domain. i.e. you can plug any number for x, and this will not make a function undefined:

Calculus wise, at any point, (for any x=a)
1) \(\large\color{midnightblue}{ f(a) }\) exisits

2) \(\large\color{midnightblue}{ \displaystyle\lim_{x \rightarrow a}f(x) }\) exist (saying that both sides of the limit are equivalent)

3) \(\large\color{midnightblue}{ \displaystyle\lim_{x \rightarrow a}f(x)=f(a) }\)

domain: \(\large\color{midnightblue}{ (-\infty,\infty)}\)

the range, gives us this:
\(\large\color{midnightblue}{(x+20)^2}\) is always positive. so the least it can get, is when x=0.
when x=-20.
when x=-20, then (x+20)^2 is zero, but it won't be a negative, so:

range: \(\large\color{midnightblue}{ [0,-\infty) }\)

Answer 2

assuming a typo, it should be: \(\large\color{midnightblue}{ y=(x-2)^2+2 }\)
for x it is a continous function, and y will be at least a 2.

Answer 3

when x=0, the y=6
when y=0 the x=2\(\color{midniblue}{ \pm}\)\(\color{midniblue}{ i\sqrt{2}}\)

Answer 4

(no real number y-intercept)

Answer 5

next time I would ask one question at a time, and make no typos, you forgot the slashes...
good luck

Answer 6

thanks so much, how do you give a medal?

Answer 7

so what to do with helping you

Answer 8

I really need more help on number 2
Give the asmyptotes and intercepts of the function:
y=1/(x-2)+2