Which point is the vertex of the quadratic function f(x) = x^2-6+8? A (0,8)
B (3,-1)
C (3,8)
D (-3,-1)

Step-by-step explanation:

f(x)= x^2 - 6x + 8

a of s= 6/2 = 3

(3)^2 - 6(3) + 8

9 - 18 + 8

-9 + 8 = -1

(3, -1)

answer is B

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.

mrbilldoesmath!

answer: a parabola

steps:

(x-1)(x+4)   = x(x+4) - 1(x+4) = (x^2 + 4x) - (x+4)

                = x^2 +4x -x -4   = x^2 +3x -4

which is a parabola

mrb