How to Graph a Rational Function

A rational function is an equation that takes the form y = N(x)/D(x) where N and D are polynomials. Attempting to sketch an accurate graph of one by hand can be a comprehensive review of many of the most important high school math topics from basic algebra to differential calculus.^{[1]XResearch source} Consider the following example: y = (2x² - 6x + 5)/(4x + 2).

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^{[2]XResearch source} Simply set x = 0. Everything but the constant terms vanish, leaving y = 5/2. Expressing this as a coordinate pair, (0, 5/2) is a point on the graph. Graph that point.
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Long divide the denominator into the numerator to determine the behavior of y for large absolute values of x. In this example, division shows that y = (1/2)x - (7/4) + 17/(8x + 4). For large positive or negative values of x, 17/(8x + 4) approaches zero, and the graph approximates the line y = (1/2)x - (7/4). Using a dashed or lightly drawn line, graph this line.^{[3]XResearch source}
- If the degree of the numerator is less than the degree of the denominator, there is no division to do, and the asymptote is y = 0.
- If deg(N) = deg(D), the asymptote is a horizontal line at the ratio of the leading coefficients.
- If deg(N) = deg(D) + 1, the asymptote is a line whose slope is the ratio of the leading coefficients.
- If deg(N) > deg(D) + 1, then for large values of |x|, y quickly goes to positive or negative infinity as a quadratic, cubic, or higher degree polynomial. In this case, it probably isn't worthwhile to accurately graph the quotient of the division.
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Find the zeros. A rational function has a zero when it's numerator is zero, so set N(x) = 0. In the example, 2x² - 6x + 5 = 0. The discriminant of this quadratic is b² - 4ac = 6² - 4*2*5 = 36 - 40 = -4. Since the discriminant is negative, N(x), and consequently f(x), has no real roots. The graph never crosses the x-axis. If any zeros were found, add those points to the graph.^{[4]XResearch source}
Find the vertical asymptotes. A vertical asymptote occurs when the denominator is zero.^{[5]XResearch source} Setting 4x + 2 = 0 gives the vertical line x = -1/2. Graph each vertical asymptote with a light or dashed line. If some value of x makes both N(x) = 0 and D(x) = 0, there may or may not be a vertical asymptote there. This is rare, but see the tips for how to deal with it if it occurs.
When is it positive, negative, or zero? In the example, the numerator of the remainder is 17 which is always positive. The denominator, 4x + 2, is positive to the right of the vertical asymptote and negative to the left. This means that the graph approaches the linear asymptote from the above for large positive values of x and from below for large negative values of x. Since 17/(8x + 4) can never be zero, this graph never intersects the line y = (1/2)x - (7/4). Don't add anything to the graph right now, but note these conclusions for later.^{[6]XResearch source}
^{[7]XResearch source} A local extremum may occur whenever N'(x)D(x)- N(x)D'(x) = 0. In the example, N'(x) = 4x - 6 and D'(x) = 4. N'(x)D(x) - N(x)D'(x) = (4x - 6)(4x + 2) - (2x² - 6x + 5)*4 = 0. Expanding, combining terms, and dividing by 4 leaves x² + x - 4 = 0. The quadratic formula shows roots near x = 3/2 and x = -5/2. (These differ by about 0.06 from the exact values, but our graph isn't going to be precise enough to worry about that level of detail. Choosing a decent rational approximation makes the next step easier.)
^{[8]XResearch source} Plug the x-values from the previous step back into the original rational function to find the corresponding y-values. In the example, f(3/2) = 1/16 and f(-5/2) = -65/16. Add these points, (3/2, 1/16) and (-5/2, -65/16), to the graph. Since we approximated in the previous step, these aren't the exact minima and maxima, but are probably close. (We know (3/2, 1/16) is very close to the local minimum. From step 3, we know that y is always positive when x > -1/2 and we found a value as small as 1/16, so at least in this case, the error is probably less than the thickness of the line.)
and smoothly extend the graph from the known points to the asymptotes taking care to approach them from the correct direction.^{[9]XResearch source} Take care not to cross the x-axis except at the points already found in step 3. Don't cross the horizontal or linear asymptote except at the points already found in step 5. Don't change from upward sloping to downward sloping except at the extreme found in the previous step.^{[10]XResearch source}

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How do I create a graph has no x intercept?
Donagan
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That would be a graph of a function where y is never equal to zero. An example is y = x² + 1. There is no x value for which the corresponding y value is zero. The graph is a parabola opening upward from a minimum y value of 1.
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Don't we at some point take the Limit of the function? If not then, on what kind of the function can we do that?
Orangejews
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The step about horizontal asymptotes finds the limit as x goes to + and - infinity. You might also take one-sided limits at each vertical asymptote to see if the graph approaches +inf or -inf from each side.
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Some of these steps may involve solving a high degree polynomial.^{[11]XResearch source} If you cannot find exact solutions through factorization, formulae, or other means, then estimate the solutions using numerical techniques such as Newton's method.
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If you follow the steps in order it usually isn't necessary to use second derivative tests or similar potentially complicated methods to determine if the critical values are local maxima, local minima, or neither. Try to use the information from previous steps and a little logic first.
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If you are trying to do this with only precalculus methods, you can replace the steps about finding the local extrema by computing several additional (x, y) ordered pairs between each pair of asymptotes. Alternatively, if you don't care why it works, there's no reason why a precalculus student can't take the derivative of a polynomial and solve N'(x)D(x) - N(x)D'(x) = 0.
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