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Graphing Quadratic Functions: Examples (page 4 of 4) Sections: Introduction, The meaning of the leading coefficient / The vertex, Examples
Since it is so simple to find the yintercept (and it will probably be a point in my Tchart anyway), they are only asking for the xintercepts this time. To find the xintercept, I set y equal 0 and solve:
0 = –x^{2} – 4x + 2 x^{2} + 4x – 2 = 0 For graphing purposes,
the intercepts are at about (–4.4,
0) and (0.4,
0). (When I write
down the answer, I will of course use the "exact" form, with
the square roots; my calculator's decimal approximations are just for
helping me graph.) To find the vertex, I look at the coefficients: a = –1 and b = –4. Then: h = ^{–(–4)}/_{2(–1)} = –2 To find k, I plug h = –2 in for x in y = –x^{2} – 4x + 2, and simplify: k = –(–2)^{2} – 4(–2) + 2 = –4 + 8 + 2 = 10 – 4 = 6 Now I'll find some additional plot points, to help me fill in my graph: Note that I picked xvalues that were centered around the xcoordinate of the vertex. Now I'll plot the parabola: The vertex is at (–2, 6), and the intercepts are at the following points: (0, 2), , and
To find the vertex, I look at the coefficients: a = –1 and b = 2. Then: h = ^{–(2)}/_{2(–1)} = 1 To find k, I'll plug h in for x and simplify: k = –(1)^{2} + 2(1) – 4 = –1 + 2 – 4 = 2 – 5 = –3 The vertex is below the xaxis, and, since this is a negative quadratic, I know that the parabola is going to be upsidedown. So can my line possibly cross the xaxis? Can there possibly be any xintercepts? Of course not! So I expect to get "no (real) solution" when I try to find the xintercepts, but I need to show my work anyway. To find the xintercept, I set y equal 0 and solve: 0 = –x^{2} + 2x – 4 x^{2} – 2x + 4 = 0 As soon as I get a negative inside the square root, I know that I can't get a graphable solution. So, as expected, there are no xintercepts. Now I'll find some additional plot points, to fill in my graph: Note that I picked xvalues that were centered around the xcoordinate of the vertex. Now I'll plot the parabola: Copyright © Elizabeth Stapel 20022011 All Rights Reserved The vertex is at (1, –3), and the only intercept is at (0, –4). This last exercise 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 xintercepts, or if you can go straight on to plotting some points and drawing the graph. If the vertex is below the xaxis (that is, if the yvalue is negative) and the quadratic is negative (so the parabola opens downward), then there will be no xintercepts. Similarly, if the vertex is above the xaxis (that is, if the yvalue is positive) and the quadratic is positive (so the parabola opens upward), then there will be no xintercepts. In most of the graphs that I did (though not the first one), it just so happened that the points on the Tchart 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 Tchart points might not be symmetric. In particular, the Tchart points will not "match" if the xcoordinate of the vertex is something other than a whole number or a halfnumber (such as "3.5"). Warning: Don't expect the plotpoints always to "match up" on either side of the vertex; in particular, don't do half the points on your Tchart and then "fill in" the rest of your Tchart 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 Tchart. You will either end up with a really tall graph or else a rather short Tchart. 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 rightsideup and draw the "sides" of the parabola. Warning: 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 standard sheet of paper, then you're drawing your graphs way too small. << Previous Top  1  2  3  4  Return to Index



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