For example, suppose you begin with the function x−25x2+5x{\displaystyle {\frac {x-2}{5x^{2}+5x}}}. The denominator 5x2+5x{\displaystyle 5x^{2}+5x} can be factored into the two terms (5x)(x+1){\displaystyle (5x)(x+1)}. As another example, consider the function y=3x+1x2+2x+1{\displaystyle y={\frac {3x+1}{x^{2}+2x+1}}}. You should recognize the denominator as a simple quadratic function, which can be factored into (x+1)(x+1){\displaystyle (x+1)(x+1)}. Recognize that some denominator functions may not be able to be factored. For example, in the equation y=x2−2x2+3x−1{\displaystyle y={\frac {x^{2}-2}{x^{2}+3x-1}}}, the function in the denominator, x2+3x−1{\displaystyle x^{2}+3x-1} cannot be factored. For this first step, you will just have to leave it in that form. If you need to review factoring of functions, check out the articles Factor Algebraic Equations or Factor Second Degree Polynomials (Quadratic Equations).
For example, if a denominator function factored as (5x)(x+1){\displaystyle (5x)(x+1)}, then you would set this equal to 0 as (5x)(x+1)=0{\displaystyle (5x)(x+1)=0}. The solutions will be any values of x that make this true. To find those values, set each individual factor equal to 0, to create two mini-problems of 5x=0{\displaystyle 5x=0} and x+1=0{\displaystyle x+1=0}. The first solution is x=0{\displaystyle x=0} and the second is x=−1{\displaystyle x=-1}. Given another example with a denominator of x2+5x+6{\displaystyle x^{2}+5x+6}, this could be factored into the two terms (x+3)(x+2){\displaystyle (x+3)(x+2)}. Setting each factor equal to 0 leads to x+3=0{\displaystyle x+3=0} and x+2=0{\displaystyle x+2=0}. Therefore, the solutions for this problem would be x=−3{\displaystyle x=-3} and x=−2{\displaystyle x=-2}.
For example, a simple equation like y=2x{\displaystyle y=2x} will have infinite solutions. Written in pairs of (x,y), some possible solutions are (1,2), (2,4), (3,6), or any pair of numbers in which the second number is double the first. Plotting these points on the x,y coordinate plane will show a continuous straight line that appears as a diagonal that goes upward from left to right. To see more samples of this type of graph, you may want to review Graph Linear Equations. A graph of a quadratic equation is one that has an exponent of 2, such as y=x2+2x−1{\displaystyle y=x^{2}+2x-1}. Some sample solutions are (-1,-2), (0,-1), (1,1), (2,7). If you plot these points, and others, you will find the graph of a parabola, which is a u-shaped curve. To review this type of graph, you can look at Graph a Quadratic Equation. If you need more help reviewing how to graph functions, read Graph a Function or Graph a Rational Function.
For example, consider the equation y=1x{\displaystyle y={\frac {1}{x}}}. If you begin at the value x=3 and count down to select some solutions for this equation, you will get solutions of (3, 1/3), (2, 1/2), and (1,1). If you continue counting down, the next value for x would be 0, but this would create the fraction y=1/0. Because division by 0 is undefined, this cannot be a solution to the function. Therefore, the value of x=0 is a vertical asymptote for this equation.
