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Solving Quadratic Equations:
    Solving "by Graphing"
(page 5 of 6)

Sections: Solving by: factoring, taking roots, completing the square, using the formula, graphing


To be honest, solving "by graphing" is an achingly trendy but somewhat bogus topic. The basic idea behind solving by graphing is that, since the "solutions" to "ax2 + bx + c = 0" are the x-intercepts of "y = ax2 + bx + c", you can look at the x-intercepts of the graph to find the solutions to the equation. There are difficulties with "solving" this way, though....

When you graph a straight line like "y = 2x + 3", you can find the x-intercept (to a certain degree of accuracy) by drawing a really neat axis system, plotting a couple points, grabbing your ruler and drawing a nice straight line, and reading the (approximate) answer from the graph with a fair degree of confidence.

On the other hand, a quadratic graphs as a wiggly parabola. If you plot a few non-x-intercept points and then draw a curvy line through them, how do you know if you got the x-intercepts even close to being correct? You don't. The only way you can be sure of your x-intercepts is to set the quadratic equal to zero and solve. But the whole point of this topic is that they don't want you to do the (exact) algebraic solving; they want you to guess from the pretty pictures.

 

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So "solving by graphing" tends to be neither "solving" nor "graphing". That is, you don't actually graph anything, and you don't actually do any of the "solving". Instead, you are told to punch some buttons on your graphing calculator and look at the pretty picture, and then you're told which other buttons to hit so the software can compute the intercepts (or you're told to guess from the pretty picture in the book, hoping that the printer lined up the different print runs for the different ink colors exactly right). I think the educators are trying to "help" you "discover" the connection between x-intercepts and solutions, but the concept tends to get lost in all the button-pushing. Okay, enough of my ranting...


To "solve" by graphing, the book may give you a very neat graph, probably with at least a few points labelled; the book will ask you to state the points on the graph that represent solutions. Otherwise, it will give you a quadratic, and you will be using your graphing calculator to find the answer. Since different calculator models have different key-sequences, I cannot give instruction on how to "use technology" to find the answers, so I will only give a couple examples of how to solve from a picture that is given to you.

  • Solve x2 – 8x + 15 = 0 by using the following graph.
    • graph of y = x^2 - 8x + 15

    The graph is of the related quadratic, y = x2 – 8x + 15, with the x-intercepts being where y = 0. The point here is to look at the picture (hoping that the points really do cross at whole numbers, as it appears), and read the x-intercepts (and hence the solutions) from the picture.

      The solution is x = 3, 5 Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

Since x2 – 8x + 15 factors as (x – 3)(x – 5), we know that our answer is correct.

  • Solve 0.3x2 – 0.5x –  5/3 = 0 by using the following graph.
    • graph of y = 0.3x^2 - 0.5x - 5/3

    For this picture, they labelled a bunch of points. Partly, this was to be helpful, because the x-intercepts are messy (so I could not have guessed their values without the labels), but mostly this was in hopes of confusing me, in case I had forgotten that only the x-intercepts, not the vertices or y-intercepts, correspond to "solutions".

    The x-values of the two points where the graph crosses the x-axis are the solutions to the equation.

      The solution is  x = –5/3, 10/3

  • Find the solutions to the following quadratic:
    • graph of quadratic

    They haven't given me the quadratic equation, so I can't check my work algebraically. (And, technically, they haven't even given me a quadratic to solve; they have only given me the picture of a parabola from which I am supposed to approximate the x-intercepts, which really is a different question....)

    I ignore the vertex and the y-intercept, and pay attention only to the x-intercepts. The "solutions" are the x-values of the points where the pictured line crosses the x-axis:

      The solution is  x = –5.39, 2.76


"Solving" quadratics by graphing is silly in "real life", and requires that the solutions be the simple factoring-type solutions such as "x = 3", rather than something like "x = –4 + sqrt(7)". In other words, they either have to "give" you the answers (by labelling the graph), or they have to ask you for solutions that you could have found easily by factoring. About the only thing you can gain from this topic is reinforcing your understanding of the connection between solutions and x-intercepts: the solutions to "(some polynomial) equals (zero)" correspond to the x-intercepts of "y equals (that same polynomial)". If you come away with an understanding of that concept, then you will know when best to use your graphing calculator or other graphing software to help you solve general polynomials; namely, when they aren't factorable.

But in practice, given a quadratic to solve, you should not start by drawing a graph. Which raises the question: For any given quadratic, which method should you use?

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Cite this article as:

Stapel, Elizabeth. "Solving Quadratic Equations: Solving by Graphing." Purplemath. Available from
    http://www.purplemath.com/modules/solvquad5.htm. Accessed
 

 

 

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