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Graphing Quadratic Functions: Examples (page 3 of 4) Sections: Introduction, The meaning of the leading coefficient / The vertex, Examples
This is the same quadratic as in the last example. I already found the vertex when I worked the problem above. This time, I also need to find the intercepts before I do my graph. To find the y-intercept, I set x equal to zero and solve: y = 3(0)2 + (0) – 2 = 0 + 0 – 2 = –2 Then the y-intercept is the point (0, –2). To find the x-intercept, I set y equal to zero, and solve: 0 = 3x2 + x
– 2 Then the x-intercepts are at the points (–1, 0) and ( 2/3, 0).
To find the y-intercept, I set x equal to 0 and solve: y = (0)2 – (0) – 12 = 0 – 0 – 12 = –12 To find the x-intercept, I set y equal to 0 and solve: 0 = x2 – x – 12 0 = (x – 4)(x + 3) x = 4 or x = –3 To find the vertex, I look at the coefficients: a = 1 and b = –1. Plugging into the formula, I get: h = –(–1)/2(1) = 1/2 = 0.5 To find k, I plug h = 1/2 in for x inside y = x2 – x – 12, and simplify: k = (1/2)2 – (1/2) – 12 = 1/4 – 1/2 – 12 = –12.25 Once I have the vertex, it's easy to write down the axis of symmetry: x = 0.5. Now I'll find some additional plot points, to fill in the graph:
For my own convenience, I picked x-values that were centered around the x-coordinate of the vertex. Now I can plot the parabola: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved
The vertex is at the point (0.5, –12.25), << Previous Top | 1 | 2 | 3 | 4 | Return to Index Next >>
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