Using long division, you can calculate that f(x)=( 2x^3 + 14x^2 + 7x - 29 ) / (2x^2 + 5 ) = Q(x) + ( R(x) / (2x^2 +5) )

The quotient Q(x) is

The remainder R(x) is

From this you can conclude that y=f(x) has an oblique (or slant) asymptote given by the line with equation y = R(x) or Q(x)

because lim as x approaches positive or negative infinity (R(x)) / (2x^2+5) = .

The Quotient Q(x) is (x + 7)

The Remainder R(x) is (2x - 64)

Limit as x approaches negative or positive infinity (R(x)/(2x² + 5)) is 0

Step-by-step explanation:

We want to use long division to determine the remainder and quotient in

(2x³+ 14x² + 7x - 29) / (2x² + 5)

This would be done step by step, as follows:

The denominator is written outside, and the numerator inside

We look for a number the when multiplied by the first expression of the funtion outside, gives the first expression of the funtion inside.

We multiply this function by the function outside, and write the resulting values under and subtract from the function inside.

The result is written, and the process is repeated, until the highest power of the function outside is greater the highest power of the function inside.

NOW, LET'S DO IT!

x times 2x = 2x³, so we choose x.

x

2x² + 5 | 2x³+ 14x² + 7x - 29

| 2x³ + 5x

| 14x² + 2x - 29

7 times 2x² gives 14x², we choose 7

x + 7

2x² + 5 | 2x³+ 14x² + 7x - 29

| 2x³ + 5x

| 14x² + 2x - 29

| 14x² + 35

| 2x - 64

We can't go further as the power of x inside is smaller than the power of x outside.

Therefore

Quotient Q(x) = x + 7

Remainder R(x) = 2x - 64

Limit as x approaches negative or positive infinity (R(x)/(2x² + 5))

= lim(x->±infty) (2x - 64) / (2x² + 5)

= 2lim(x->±infty) x(1 - 32/x) / x²(2 + 5/x²)

= 2(0)

= 0

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.