Find the area enclosed in the graph of
x² + y² 16x + 32y.

3

256

sq.units

Step-by-step explanation:

Both parabolas cut each other at (0,0) and (16,16)

Area enclosed by these parabolas

=∫

0

16

4

x

dx−∫

0

16

16

x

2

dx

=[

3

2×4×x

3/2

]

0

16

−[

16×3

x

3

]

0

16

=

3

2×4

4

−

3

4

4

=

3

256

sq. units

where are the choices?

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.