Print-friendly page

Graphing Quadratic Functions: Examples (page 4 of 4)

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

• Find the x-intercepts and vertex of y = –x² – 4x + 2.

Note the difference here: since it is so simple to find the y-intercept (and it will probably be a point in my T-chart anyway), they are only asking for the x-intercepts this time. To find the x-intercept, set y equal 0 and solve:

0 = –x² – 4x + 2

x² + 4x – 2 = 0



For graphing purposed, the intercepts are at about (–4.4, 0) and (0.4, 0). (When you write down the answer, use the "exact" form, with the square roots; the decimal approximations are just for helping you graph.)

To find the vertex, look at the coefficients: a = –1 and b = –4. Then:

h = ^–(–4)/_2(–1) = –2

To find k, plug h in for x and simplify:

k = –(–2)² – 4(–2) + 2 = –4 + 8 + 2 = 10 – 4 = 6

Now 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 plot the parabola:

The vertex is at (–2, 6), and the intercepts are at the following points:

(0, 2),  , and

• Find the x-intercepts and vertex of y = –x² + 2x – 4.

To find the vertex, look at the coefficients: a = –1 and b = 2. Then:

h = ^–(2)/_2(–1) = 1

To find k, plug h in for x and simplify:

k = –(1)² + 2(1) – 4 = –1 + 2 – 4 = 2 – 5 = –3

The vertex is below the x-axis, and the parabola is upside-down. Can the line possibly cross the x-axis then? Can there possibly be x-intercepts? Of course not! So I expect to get "no (real) solution" when I try to find the x-intercepts. To find the x-intercept, set y equal 0 and solve:

0 = –x² + 2x – 4

x² – 2x + 4 = 0



As soon as you get a negative inside the square root, you know that you can't get a graphable solution. So, as expected, there are no x-intercepts. Now 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 plot the parabola:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

The vertex is at (1, –3), and the only intercept is at (0, –4).

This last problem illustrates one way you can cut down a bit on your work. If you solve for the vertex first, then you can easily tell if you need to continue on and look for the x-intercepts, or if you can go straight on to plotting some points and drawing the graph. If the vertex is below the  x-axis (that is, if the  y-value is negative) and the quadratic is negative (so the parabola opens downward), then there will be no x-intercepts. Similarly, if the vertex is above the x-axis (that is, if the y-value is positive) and the quadratic is positive (so the parabola opens upward), then there will be no x-intercepts.

In most of the graphs that I did (though not the first one), it just so happened that the points on the T-chart were symmetric about the vertex; that is, that the points "matched" on either side of the vertex. While a parabola is always symmetric about the vertical line through the vertex (the parabola's "axis"), the T-chart points might not be symmetric. In particular, the T-chart points will not "match" if the x-coordinate of the vertex is something other than a whole number or a half-number (such as "3.5"). So don't expect the points always to "match up", and don't do half the points on your T-chart and then "fill in" the rest of your T-chart by assuming a symmetry that might not exist.

Other tips for graphing: If the parabola is going to be "skinny", then expect that you will get some very large values in your T-chart. You will either end up with a really tall graph or else a rather short T-chart. If the parabola is going to be "fat", then expect that you will probably have to plot points with fractions as coordinates. In either case, when you go to connect the dots to draw the parabola, you might find it helpful to turn the paper sideways and first draw the really curvy part through the vertex, making sure that it looks nice and round. Then turn the paper back up and draw the "sides" of the parabola. By the way, you want to draw your graphs big enough to be clearly seen by your instructor; if you're fitting more than two or maybe three graphs on one side of a sheet of paper, then you're drawing your graphs way too small.

<< Previous  Top  |  1 | 2 | 3 | 4  |  Return to Index

  Copyright © 2006-2008  Elizabeth Stapel   |   About   |   Terms of Use

 Feedback   |   Error?