On a coordinate plane, a parabola opens down. It goes through (negative 2, negative 2), has a vertex at (1.5, 10), and goes through (5, negative 2).
Over which interval is the graph of f(x) = –x2 + 3x + 8 increasing?

(–∞, 1.5)
(–∞, 10.25)
(1.5, ∞)
(10.25, ∞)

a is the right answer

75%

step-by-step explanation:

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.