Normal Method of Making a Parabola
- Examining the leading coefficient of the quadratic equation gives an idea of what the parabola will look like. If the absolute value of "a" is greater than 1, the parabola will have a narrow opening. If "a" is negative, the parabola will open down. If "a" is positive, the parabola will open up.
For example, looking at the quadratic equation "y = 5x^2 + 11x + 2" allows the prediction that, based on a leading coefficient of 2, the parabola will be have a wider opening and face up. - The first point found for the parabola is the vertex, which forms the lowest point of upward-facing parabolas or the highest point of down-facing parabolas. For a vertex point of (h, k), the quadratic equation can be rewritten in the form "y = a(x - h)^2 + k." But a simplified vertex formula is "h = -b / 2a," which is solved then plugged back in as "x" in the standard form of the equation then solved for "k."
For example, using again the equation "y = 5x^2 + 11x + 2 , h = -11 / (2 * 5) = (-11/10) = -1.1." Plug the value back into the equation as "x": "5((-1.1)^2) + 11(-1.1) + 2 = 5(1.21) + -12.1 + 1 = 6.05 + -12.1 + 1 = -5.05." The vertex is the point (-1.1, -5.05). - Find the y-intercept of the parabola by setting the "x" of the standard form equal to 0 then solving. Find the x-intercepts by factoring the equation and setting each of the multiples each to 0 then solving for "x".
Continuing with the example equation of "y = 5x^2 + 11x + 2," set "x" equal to zero: "y = 5(0^2) + 11(0) + 2 = 2." The y-intercept is point (0, 2). Set the "y" equal to zero and factor the equation: "0 = (5x + 1)(x + 2)." Set the first multiple equal to zero and solve: "5x + 1 = 0," subtract 1 from both sides for "5x = -1," then divide both by 5 for "x = (-1/5)." Set the second set equal to zero and solve; "(x + 2) = 0," subtract 2 for "x = -2." Thus the x-intercepts are points (-1/5, 0) and (-2, 0). - Create a t-chart to find additional points for the parabola. Draw a t-shape consisting of two columns with two headers. Label the left column "x" and the right with the equation, such as the example equation "5x^2 + 11x + 2." Write four test values for "x" in the left column to substitute into the equation.
For this example, find the values when "x" equals -3, -1, 1 and 2 (note that 0 was found already with the y-intercept and -2 was one of the x-intercepts). Plug -3 into the equation to solve: "5(-3^2) + 11(-3) + 2 = 45 + -33 + 2 = 14" or point (-3, 14). Plug -1 into the equation: "5(-1^2) + 11(-1) + 2 = 5 + -11 + 2 = -4" or point (-1, -4). Plug 1 into the equation: "5(1^2) + 11(1) + 2 = 5 + 11 + 2 = 18" or point (1, 18). Plug 2 into the equation: "5(2^2) + 11(2) + 2 = 20 + 22 + 2 = 44" or point (2, 44). - Graph on a rectangular coordinate system the vertex, intercepts and t-chart points. If you need more points to see the shape of the parabola, return to the t-chart and solve two or three more values of "x." Connect the lines to form the parabola then draw arrows at the ends of the extending lines to indicate continuation.
