Find vertical and horizontal asymptotes of the function fx 1

Find vertical and horizontal asymptotes of the function: f(x)= 1/x - 8 Find the standard equation of any parabola that has vertex V(-6, 7). Find the minimum value of the function f(x) = x^2 + 6x +16 Find all values of x such that f(x)

Solution

11.) f(x) = 1/ (x-8)

vertical asymptotes set denomintor = 0

x - 8 = 0

add 8 on both sides

x = 8 is vertical asymptote

when degree of numerator is less then degree of denominator then horizontal asymptote is

y = 0 is horizontal asymptote

12) vertex = ( -6,7 )

standard equation of a parabola in vertex form is given by

y = a (x - h)^2 + k

where h,k are the vertex

therefore plugging the values we get

y = a ( x+6)^2 + 7

or

y = (x+6)^2 + 7

y = x^2 + 12x + 43

13) f(x) = x^2 + 6x + 16

minimum value occurs at

x = -b/2a

x = -6/2= -3

minimum value is

f(-3) = 7