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Graphing Quadratic Functions: Examples (page 3 of 4)

Sections: Introduction, The meaning of the leading coefficient / The vertex, Examples

• Find the vertex and intercepts of y = 3x² + x – 2 and graph; remember to label the vertex.

This is the same quadratic as in the last example. I already found the vertex when I worked the problem above. The new part here is that 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
0 = (3x – 2)(x + 1)
3x – 2 = 0  or   x + 1 = 0
x =  ²/₃   or  x = – 1

Then the x-intercepts are at the points (–1, 0) and ( ²/₃, 0).

The complete answer is a listing of the vertex and the intercepts, along with a neat graph:       The vertex is at ( –1/6 , –25/12 ), and the intercepts are at (0, –2), (–1, 0), and ( 2/3, 0).

• Find the intercepts, the axis of symmetry, and vertex of y = x² – x – 12.

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 in for x 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:

Note that I picked x-values that were centered around the x-coordinate of the vertex. Now I can plot the parabola:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

The vertex is at the point (0.5, –12.25),
   the axis of symmetry is the line x = 0.5,
      and the intercepts are at the points (0, –12),  (–3, 0), and (4, 0).

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