How to Graph an Equation Using the Point Plotting Method
- 1). Set the equation to 0 and solve for "x" to find the x-intercept(s). For example, setting the equation x^2 + 2x + 1 to 0 finds: 0 = x^2 + 2x + 1 = (x + 1)(x + 1). Now, the right-sided expression equals zero when x = -1. So, the x-intercept for this equation is at (-1, 0). Plot the point on the graph at the point.
- 2). Set the "x" variable to zero and solve for "y" to obtain the y-intercept(s). For example, setting x = 0 in the equation x^2 + 2x + 1 finds: y = 0^2 + 2(0) + 1 = 1. So, the y-intercept for this equation is at (0, 1). Plot the point on the graph at that point.
- 3). Substitute several x-coordinate points into the original equation and solve to find the y-coordinate points at these values. Choose points to the right and left of the x-intercept on an interval including the y-intercept. For example, substituting x-coordinates x = -4, x = -3, x = -2, x = 0, x = 1, x = 2 and x = 3 finds: y(-4) = -4^2 + 2(-4) + 1 = 9, y(-3) = -3^2 + 2(-3) + 1 = 4, y(-2) = -2^2 + 2(-2) + 1 = 3, y(-1) = -1^2 + 2(-1) + 1 = 0, y(0) = 0^2 + 2(0) + 1 = 1, y(1) = 1^2 + 2(1) + 1 = 4, y(2) = 2^2 + 2(2) + 1 = 9, y(3) = 3^2 + 2(3) + 1 = 16.
- 4). Plot the points on the graph. For example, since it was found that y(-4) = -4^2 + 2(-4) + 1 = 9, y(-3) = -3^2 + 2(-3) + 1 = 4, y(-2) = -2^2 + 2(-2) + 1 = 3, y(-1) = -1^2 + 2(-1) + 1 = 0, y(0) = 0^2 + 2(0) + 1 = 1, y(1) = 1^2 + 2(1) + 1 = 4, y(2) = 2^2 + 2(2) + 1 = 9, y(3) = 3^2 + 2(3) + 1 = 16, for y = x^2 + 2x + 1, the points to be plotted are: (-4, 9), (-3, 4), (-2, 3), (-1, 0), (0, 1), (1, 4), (2, 9) and (3, 16).
- 5). Draw a smooth curve connecting each of the points together, moving from the leftmost point to the right.
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