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Graphing Quadratic Functions: The
     Leading Coefficient / The Vertex (page 2 of 4)

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

The general form of a quadratic is "y = ax² + bx + c". For graphing, the leading coefficient "a" indicates how "fat" or "skinny" the parabola will be.

For | a | > 1 (such as a = 3 or a = –4), the parabola will be "skinny", because it grows more quickly (three times as fast or four times as fast, respectively, in the case of our sample values of a).  For
| a | < 1 (such as a = ¹/₃ or a = ^–1/₄ ), the parabola will be "fat", because it grows more slowly (one-third as fast or one-fourth as fast, respectively, in the examples). Also, if a is negative, then the parabola is upside-down.

As you can see, as the leading coefficient goes from way negative to slightly negative to zero (not really a quadratic) to slightly positive to way positive, the parabola goes from skinny upside-down to fat upside-down to a straight line (called a "degenerate" parabola) to a fat right-side-up to a skinny right-side-up.   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

There is a simple, if slightly "dumb", way to remember the difference between right-side-up parabolas and upside-down parabolas:

positive quadratic y = x2	negative quadratic y = –x2

One point for you to remember is that, if you have an equation where a is, say, negative, and you're coming up with plot points that make it look like the quadratic is right-side-up, then you need to go back and check your work.

Parabolas always have a lowest point (or a highest point, if the parabola is upside-down). This point, where the parabola changes direction, is called the "vertex".

If the quadratic is written in the form y = a(x – h)² + k, then the vertex is the point (h, k). This makes sense, if you think about it. The squared part is always positive (for a right-side-up parabola), unless it's zero. That is, you'll always have the fixed value k, and then you're always adding something to it to make y bigger, unless of course the squared part is zero. So the smallest y can be is k, and this smallest value will happen when the squared part equals zero. And the squared part is zero when x – h = 0, or when x = h. The same reasoning works, with k being the largest value and the squared part always subtracting from it, for upside-down parabolas.

(The "a" in the vertex form "y = a(x – h)² + k" of the quadratic is the same as the "a" in "y = ax² + bx + c".)

So you can see where the vertex form of the quadratic might be useful for finding the vertex, which is a rather distinctive point on the graph, especially if the vertex isn't one of your T-chart values. However, quadratics are not usually written in vertex form. You can complete the square to convert ax² + bx + c to vertex form, but it's simpler to just use a formula (derived from the completing-the-square process) to find the vertex.

For a given quadratic y = ax² + bx + c, the vertex (h, k) is found by computing h = ^–b/_2a, and then evaluating y at h to find k. If you've already learned the Quadratic Formula, you may find it easy to memorize the formula for k, since it is related to both the formula for h and the discriminant in the Quadratic Formula:  k = (4ac – b²) / 4a. It can be useful to be able to find the vertex, since the vertex is not always a nice neat point on your table of values. For instance:

• Find the vertex of y = 3x² + x – 2 and graph.

To find the vertex, look at the coefficients a, b, and c. The formula for the vertex gives:

h = ^–b/_2a = ^–(1)/₂₍₃₎ = ^–1/₆

Then I can find k by evaluating y at h:

k = 3(^–1/₆)² + (^–1/₆) – 2

= ³/₃₆   –  ¹/₆  – 2

=  ¹/₁₂   –  ²/₁₂  –  ²⁴/₁₂

=  ^–25/₁₂

So now I know that the vertex is at ( ^–1/₆ , ^–25/₁₂ ). Note that this point is not one that I was likely to get on my T-chart.

I need additional points for my graph:

Now I can do my graph, and I will label the vertex:

When you write down the vertex in your homework, write down "( ^–1/₆ , ^–25/₁₂ )". But for graphing purposes, the decimal approximation of "(–0.2, –2.1)" may be more helpful, since it's easier to locate on the graph.

Note the points on the above graph where the line crosses the axes. The points where the parabola crosses the axes are called the "intercepts". The y-intercept occurs where x = 0 (at the point (0, –2)) and that the left-hand x-intercept occurs at (–1, 0) (that is, at a point where y = 0). This is how you find the intercepts: whichever intercept you're trying to find, set the other variable equal to zero, and solve.

The only other consideration regarding the vertex is the "axis of symmetry". If you look at a parabola, you'll notice that you could draw a vertical line right up through the middle which would split the parabola into two mirrored halves. This vertical line, right through the vertex, is called the axis of symmetry. If you're asked for the axis, write down the line "x = h", where h is just the x-coordinate of the vertex. So in the example above, the axis would be the vertical line x = ^–1/₆.

Helpful note: If your quadratic's x-intercepts happen to be nice neat numbers (so they're relatively easy to work with), a shortcut for finding the line of symmetry is to note that this line is always exactly between the two intercepts. So you can just average the two intercepts to get the location of the axis of symmetry and the x-part of the vertex. However, if you have messy x-intercepts (as in the example above) or if the quadratic doesn't actually cross the x-axis (as you'll see on the next page), then you'll need to use the formula to find the vertex.

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