A line passes through (6, 3), (8, 4), and (n, -2). Find the value of n.

n = -4

because the line (d) passes through (6;3) and (8;4)

=> (d): passes through (6;3)

          the normal vector n(1; -2)

=> (d): x - 6 - 2(y - 3) = 0

=> (d): x - 2y = 0

because d also passes through (n; -2)

=> n - 2.(-2) = 0

⇔ n + 4 = 0

⇔ n = -4

Step-by-step explanation:

step-by-step explanation:

your mom

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.