Slant Asymptotes by Long Division

Divide, and a line falls out of the quotient.

No horizontal line fits

Take y = (x² + 3x + 5)/(x + 1). The numerator has degree 2 and the denominator has degree 1. When the numerator has the higher degree the values grow without bound, so there is no horizontal asymptote: at x = 100, y = 10305/101, which is about 102.03, and at x = 1000 it is about 1002.003.

Those values are very close to x + 2: 102 at x = 100, and 1002 at x = 1000. The curve is not settling on a height, but it does seem to be settling on a line. Long division finds that line exactly.

Divide the numerator by the denominator

Divide x² + 3x + 5 by x + 1 exactly as in Long Division of Polynomials. Divide the leading terms: x² / x = x, so the quotient starts with x. Multiply back: x(x + 1) = x² + x. Subtract: (x² + 3x + 5) − (x² + x) = 2x + 5.

Run the loop again on 2x + 5. Divide: 2x / x = 2, so the quotient gains + 2. Multiply back: 2(x + 1) = 2x + 2. Subtract: (2x + 5) − (2x + 2) = 3. The 3 has no x in it, a lower degree than x + 1, so the division stops. The quotient is x + 2 and the remainder is 3.

So x² + 3x + 5 = (x + 1)(x + 2) + 3. Check: (x + 1)(x + 2) = x² + 3x + 2, and adding 3 gives x² + 3x + 5. Dividing both sides by x + 1 splits the function into a line and a fraction: y = x + 2 + 3/(x + 1).

The remainder term fades

The fraction 3/(x + 1) is the remainder over the divisor, called the remainder term. Its numerator stays 3 while its denominator grows in size, so far from the origin it shrinks toward 0: at x = 99 it is 3/100 = 0.03, and at x = 999 it is 0.003.

So far out, y is the line x + 2 plus a piece that is almost nothing. The graph runs closer and closer to the straight line y = x + 2. A line like this, which is neither horizontal nor vertical, is called a slant asymptote, or an oblique asymptote. The slant asymptote is the quotient; the remainder only says how far the curve is from it.

−101−11999x + 2−99−911101y−99.03−9.311.3101.03

Values of y = (x² + 3x + 5)/(x + 1) beside the heights of the line y = x + 2. The gap is 3/(x + 1): 0.3 above the line at x = 9, 0.03 above at x = 99, and the same sizes below the line at x = −11 and x = −101.

Which side the curve is on

The sign of the remainder term says which side of the line the curve is on. For x greater than −1, x + 1 is positive, so 3/(x + 1) is positive and the curve is above the line. For x less than −1, x + 1 is negative, so 3/(x + 1) is negative and the curve is below the line.

The curve never meets the line: that would need 3/(x + 1) = 0, and a fraction with numerator 3 is never zero. So the curve closes onto y = x + 2 from above on the right and from below on the left. The denominator is zero at x = −1, so the line x = −1 is a vertical asymptote.

xy

The graph of y = (x² + 3x + 5)/(x + 1), with its asymptotes dashed: the slant line y = x + 2 and the vertical line x = −1. Right of x = −1 the curve is above the slant line, and left of it the curve is below.

A negative remainder

Take y = (x² + x − 5)/(x − 1). Divide: x² / x = x, and x(x − 1) = x² − x. Subtract: (x² + x − 5) − (x² − x) = 2x − 5. Then 2x / x = 2, and 2(x − 1) = 2x − 2. Subtract: (2x − 5) − (2x − 2) = −3. So y = x + 2 − 3/(x − 1), and the slant asymptote is again y = x + 2.

This time the remainder term is −3/(x − 1). For x greater than 1 it is negative, so the curve is below the line; at x = 4, y = 15/3 = 5, while the line gives 6. For x less than 1 it is positive, so the curve is above the line; at x = −2, y = (−3)/(−3) = 1, while the line gives 0. The vertical asymptote is x = 1.

xy

The graph of y = (x² + x − 5)/(x − 1), with its asymptotes dashed at y = x + 2 and x = 1. The point (4, 5) is below the slant line and the point (−2, 1) is above it: the sides are the other way round from the first example.

Other degrees

What gives a slant asymptote is the difference between the degrees, not the degrees themselves: the top must be exactly one degree higher than the bottom. In y = (x³ + 2)/x², the top has degree 3 and the bottom degree 2. Dividing each term of the top by x² gives y = x + 2/x², so the slant asymptote is y = x. Here the remainder term 2/x² is positive for every x other than 0, so the curve is above the line on both sides.

If the top is two or more degrees higher, the quotient is not a line. In y = (x³ + 1)/x = x² + 1/x, the quotient is x², so far out the curve runs close to the parabola y = x², and there is no slant asymptote.

The usual mistakes

Keeping the whole middle coefficient. For (x² + 3x + 5)/(x + 1), y = x + 3 is not the asymptote: the first round of the division already used x of the 3x, leaving 2x. The quotient is x + 2.

Stopping after the first term. y = x is only the start of the quotient; the loop runs again on 2x + 5, and the slant asymptote is y = x + 2.

Looking for a horizontal asymptote from the leading coefficients. That rule is for a top and a bottom of the same degree. Here the top is one degree higher, so the curve follows a slanted line instead of settling on a height.

Taking the remainder as part of the asymptote. The asymptote is the quotient alone. The remainder term 3/(x + 1) is the gap between the curve and the line, and it shrinks toward 0.

Worked example: The Average Cost of a Chair When Overtime Makes Large Outputs Expensive: A Slant Asymptote

Question A workshop makes x chairs in a day. Wood, wages and overtime bring the total cost of the day to x2 + 50x + 400 dollars, so the average cost of a chair is A(x) = x2 + 50x + 400x dollars. (a) Divide to write A(x) as a quotient plus a remainder term, and give the slant asymptote of the graph. Say what the asymptote tells the workshop. (b) Find the two outputs at which the average cost is $100 a chair.

  1. 1.Divide each term of the top by x: x2x = x, 50xx = 50, and 400 is left over as the remainder. So A(x) = x + 50 + 400x. The quotient is x + 50 and the remainder term is 400x.

    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)divide each term of the top by xA(x) = x + 50 + 400/x
    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)divide each term of the top by xA(x) = x + 50 + 400/x
    Divide each term of the top by x: A(x) = x + 50 + 400x. The quotient is x + 50, and the remainder term is 400x.

  2. 2.As x grows, the remainder term 400x fades toward 0, and the curve closes on the line y = x + 50. For example A(400) = 400 + 50 + 1 = 451, and the line gives 450.

    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)400/x fades as x growsA(400) = 451, and the line y = x + 50 gives 450
    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)400/x fades as x growsA(400) = 451, and the line y = x + 50 gives 450
    The remainder term 400x fades as x grows, so the curve closes on the dashed line y = x + 50 from above.

  3. 3.(a) A(x) = x + 50 + 400x, and the slant asymptote is y = x + 50. For large outputs the average cost follows this line: it rises by about $1 for every extra chair, and it is always a little above the line, by 400x dollars.

    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)y = x + 50slant asymptote y = x + 50about $1 more for every extra chair
    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)y = x + 50slant asymptote y = x + 50about $1 more for every extra chair
    (a) The slant asymptote is y = x + 50. For large outputs the average cost follows it, rising by about $1 for every extra chair.

  4. 4.Put A(x) = 100: x + 50 + 400x = 100. Multiply both sides by x: x2 + 50x + 400 = 100x, so x2 − 50x + 400 = 0.

    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)y = x + 50x + 50 + 400/x = 100: multiply by xx2− 50x + 400 = 0
    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)y = x + 50x + 50 + 400/x = 100: multiply by xx2− 50x + 400 = 0
    Put A(x) = 100 and multiply both sides by x: x2 + 50x + 400 = 100x, so x2 − 50x + 400 = 0.

  5. 5.Factorize: (x − 10)(x − 40) = 0, so x = 10 or x = 40. Both are positive whole numbers of chairs, so both fit the situation.

    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)y = x + 50(x − 10)(x − 40) = 0x = 10 or x = 40
    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)y = x + 50(x − 10)(x − 40) = 0x = 10 or x = 40
    (x − 10)(x − 40) = 0, so the curve meets the level 100 at x = 10 and again at x = 40.

  6. 6.(b) The average cost is $100 at 10 chairs and at 40 chairs. Check: A(10) = 10 + 50 + 40 = 100 and A(40) = 40 + 50 + 10 = 100. Between the two outputs the average cost is lower: A(20) = 20 + 50 + 20 = 90.

    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)y = x + 50$100 a chair at 10 chairs and at 40 chairslower in between: A(20) = 20 + 50 + 20 = 90
    0501001500102030405060chairs made in a day, xaverage cost of a chair ($)y = x + 50$100 a chair at 10 chairs and at 40 chairslower in between: A(20) = 20 + 50 + 20 = 90
    (b) The average cost is $100 at 10 chairs and at 40 chairs. Between them the curve is lower, with A(20) = 90.

Answer: (a) A(x) = x + 50 + 400x, and the slant asymptote is y = x + 50: for large outputs the average cost rises by about $1 for every extra chair; (b) 10 chairs or 40 chairs

Common mistakes

  • Looking for a horizontal asymptote by comparing leading coefficients. That rule is for a top and a bottom of the same degree. Here the top is one degree higher, so the average cost does not level off: it follows the line y = x + 50, which keeps rising.
  • Stopping at x = 10. A quadratic equation can have two solutions, and the curve falls to a lowest point and rises again, so it meets the level 100 twice. The second output, 40 chairs, is just as real as the first.

More rational functions problems, worked step by step →

Worked example: Fencing a Plot That Must Hold a Lawn of Fixed Area Beside a Vegetable Bed: A Vertical and a Slant Asymptote

Question A rectangular plot is x meters wide. A vegetable bed 4 meters wide runs along the whole of one long side, and the rest of the plot is a lawn, so the lawn is x − 4 meters wide. The lawn must have an area of 48 square meters, and a fence goes round the whole plot. (a) Find a formula for the length of fence P(x) in meters, and find the vertical asymptote and the slant asymptote of its graph. Say what each one means. (b) There are 40 meters of fence. Find the two possible widths of the plot.

  1. 1.The lawn is x − 4 meters wide and has an area of 48 square meters, so the length of the lawn, and of the plot, is 48x − 4 meters. The fence goes round a plot that is x meters by 48x − 4 meters: P(x) = 2x + 96x − 4.

    the fence goes round the whole plotlawn 48 m2bedx − 44xlength 48/(x − 4)the lawn is x − 4 wide, so it is 48/(x − 4) longP(x) = 2x + 2 × 48/(x − 4) = 2x + 96/(x − 4)
    the fence goes round the whole plotlawn 48 m2bedx − 44xlength 48/(x − 4)the lawn is x − 4 wide, so it is 48/(x − 4) longP(x) = 2x + 2 × 48/(x − 4) = 2x + 96/(x − 4)
    The lawn is x − 4 meters wide and covers 48 square meters, so the plot is 48x − 4 meters long. The fence goes round the whole plot: P(x) = 2x + 96x − 4.

  2. 2.The bottom x − 4 is zero at x = 4, and the top 96 is never zero, so the vertical asymptote is x = 4. The lawn needs some width, so the domain is x > 4. Close to 4 the fence is very long: P(5) = 10 + 96 = 106.

    0204060048121620width of the plot (m), xfence (m), Px = 4x − 4 = 0 at x = 4; the top, 96, is never 0vertical asymptote x = 4; P(5) = 10 + 96 = 106
    0204060048121620width of the plot (m), xfence (m), Px = 4x − 4 = 0 at x = 4; the top, 96, is never 0vertical asymptote x = 4; P(5) = 10 + 96 = 106
    The bottom is zero at x = 4 and the top is never zero, so the vertical asymptote is x = 4 and the domain is x > 4. Close to it the fence is very long: P(5) = 106.

  3. 3.As one fraction, P(x) = 2x2 − 8x + 96x − 4. The top is one degree higher than the bottom, so there is a slant asymptote, and it is the quotient y = 2x. As x grows, the remainder term 96x − 4 fades: P(100) = 200 + 1 = 201.

    0204060048121620width of the plot (m), xfence (m), Px = 4y = 2xas one fraction: (2x2− 8x + 96)/(x − 4)the quotient is the slant asymptote: y = 2x
    0204060048121620width of the plot (m), xfence (m), Px = 4y = 2xas one fraction: (2x2− 8x + 96)/(x − 4)the quotient is the slant asymptote: y = 2x
    As one fraction, P(x) = 2x2 − 8x + 96x − 4. The top is one degree higher than the bottom, so the quotient y = 2x is a slant asymptote, and the remainder term 96x − 4 fades.

  4. 4.(a) P(x) = 2x + 96x − 4. The vertical asymptote is x = 4: a plot only just wider than the bed leaves a sliver of lawn, which must be very long to cover 48 square meters. The slant asymptote is y = 2x: a very wide plot is very short, so its fence is little more than its two long sides, 2x meters.

    0204060048121620width of the plot (m), xfence (m), Px = 4y = 2xx near 4: a sliver of lawn, which must be very longx large: the fence is little more than 2x
    0204060048121620width of the plot (m), xfence (m), Px = 4y = 2xx near 4: a sliver of lawn, which must be very longx large: the fence is little more than 2x
    (a) x = 4: a plot only just wider than the bed needs a very long sliver of lawn. y = 2x: a very wide plot is short, so its fence is little more than its two long sides.

  5. 5.Put P(x) = 40 and multiply both sides by x − 4: 2x(x − 4) + 96 = 40(x − 4), so 2x2 − 8x + 96 = 40x − 160. Collect the terms and divide by 2: x2 − 24x + 128 = 0.

    0204060048121620width of the plot (m), xfence (m), Px = 4y = 2x2x(x − 4) + 96 = 40(x − 4)x2− 24x + 128 = 0
    0204060048121620width of the plot (m), xfence (m), Px = 4y = 2x2x(x − 4) + 96 = 40(x − 4)x2− 24x + 128 = 0
    Put P(x) = 40 and multiply every term by x − 4: 2x(x − 4) + 96 = 40(x − 4), which comes to x2 − 24x + 128 = 0.

  6. 6.(b) (x − 8)(x − 16) = 0, so the plot is 8 meters or 16 meters wide. Check: at 8 meters the lawn is 4 by 12 and the plot is 8 by 12, with a perimeter of 40. At 16 meters the lawn is 12 by 4 and the plot is 16 by 4, with a perimeter of 40.

    0204060048121620width of the plot (m), xfence (m), Px = 4y = 2x(x − 8)(x − 16) = 0: 8 m or 16 m widecheck: 8 by 12 and 16 by 4 both have perimeter 40
    0204060048121620width of the plot (m), xfence (m), Px = 4y = 2x(x − 8)(x − 16) = 0: 8 m or 16 m widecheck: 8 by 12 and 16 by 4 both have perimeter 40
    (b) (x − 8)(x − 16) = 0: the plot is 8 meters or 16 meters wide. The curve meets the level 40 at both.

Answer: (a) P(x) = 2x + 96x − 4. The vertical asymptote is x = 4: a plot only just wider than the bed needs a very long sliver of lawn. The slant asymptote is y = 2x: a very wide plot is short, so its fence is little more than its two long sides; (b) 8 meters or 16 meters

Common mistakes

  • Using the width of the plot for the lawn and writing the length as 48x. The bed takes 4 meters of the width, so the lawn is x − 4 meters wide, and the length is 48x − 4. With the slip the vertical asymptote moves to x = 0 and the meaning of the bed is lost.
  • Multiplying only the fraction by x − 4 when clearing it. Every term on both sides must be multiplied: 2x becomes 2x(x − 4) and 40 becomes 40(x − 4). Leaving 2x alone gives a linear equation and only one width.

More rational functions problems, worked step by step →

Practice Slant Asymptotes by Long Division in the app