The graph shown is y= (x-2)2 - 16 = x2-4x -12. One of the things I try to get them to see is that if you break the quadratic formula into two parts, it will give you a better sense of what is happening. In the example above the quadratic formula gives \( x=\frac{4\pm \sqrt{64}}{2} = \frac{4}{2} \pm \frac{8}{2}\). I want them to see that if they break it into two fractions, the first part gives the axis of symmetry, 2, and the second gives the distance from the axis to the two real solutions, 4.
If you make the a coefficient larger, the curve will get to the x-axis sooner, and the two solutions will not be as far apart. The next image shows a quadratic with the same vertex, (2,-16) with a leading coefficient of two.
The solution then, will cross the x-axis at a distance from the axis of symmetry which is now the square root of 8 (16/2) instead of the square root of 16. As a quadratic moves c units to either side of the axis of symmetry, the quadratic will change its y-value by an amount equal to Ac<2. Setting this equal to 16 (the distance of the vertex below the x-axis) we find the distance from the axis of symmetry to the roots.
But if we look at a graph of a quadratic with complex roots, we don't see any such distance to each side of the axis of symmetry....but we can...
using a method that may have first been suggested in Howard F. Fehr, "Graphical Representation of Complex Roots," 'Multi-Sensory Aids in the Teaching of Mathematics', 'Eighteenth Yearbook of the National Council of Teachers of mathematics' [1945] pp. 130-138. George A. Yanosik, "Graphical Solutions for Complex Roots of Quadratics, Cubics, and Quartics," 'National Mathematics Magazine', 17 [Jan. 1943], pp. 147-150.]
The next graph shows a graph of a quadratic with the vertex at (3,5) and a leading coefficient of positive 2, which has no real roots.
But if we graph the quadratic with the same vertex and a leading coefficient of the opposite sign, it will cross the x-axis at a distance away from the axis that is the same as the roots of \(y=2 (x-3)^2 -5 \) . These distances imaginary coefficients of the complex solutions. AND... If you rotate the entire coordinate plane by 90o, the two points will also be the endpoints of the Argand diagram of the two solutions. I'm hoping that if a kid can fit all this together, they will begin to understand quadratic equations, their graphs, and solutions a little more. If not, we can try solving them by Newton's approach with log scales... but more about that some other blog.
The solution then, will cross the x-axis at a distance from the axis of symmetry which is now the square root of 8 (16/2) instead of the square root of 16. As a quadratic moves c units to either side of the axis of symmetry, the quadratic will change its y-value by an amount equal to Ac<2. Setting this equal to 16 (the distance of the vertex below the x-axis) we find the distance from the axis of symmetry to the roots.
But if we look at a graph of a quadratic with complex roots, we don't see any such distance to each side of the axis of symmetry....but we can...
using a method that may have first been suggested in Howard F. Fehr, "Graphical Representation of Complex Roots," 'Multi-Sensory Aids in the Teaching of Mathematics', 'Eighteenth Yearbook of the National Council of Teachers of mathematics' [1945] pp. 130-138. George A. Yanosik, "Graphical Solutions for Complex Roots of Quadratics, Cubics, and Quartics," 'National Mathematics Magazine', 17 [Jan. 1943], pp. 147-150.]
The next graph shows a graph of a quadratic with the vertex at (3,5) and a leading coefficient of positive 2, which has no real roots.But if we graph the quadratic with the same vertex and a leading coefficient of the opposite sign, it will cross the x-axis at a distance away from the axis that is the same as the roots of \(y=2 (x-3)^2 -5 \) . These distances imaginary coefficients of the complex solutions. AND... If you rotate the entire coordinate plane by 90o, the two points will also be the endpoints of the Argand diagram of the two solutions. I'm hoping that if a kid can fit all this together, they will begin to understand quadratic equations, their graphs, and solutions a little more. If not, we can try solving them by Newton's approach with log scales... but more about that some other blog.
1 comment:
This is a great post. I spent some time looking for these connections earlier this year and I found a paper describing the parabola relationship you have noted. Yanosik's paper is particularly interesting, thanks for that tip. I was able to find the paper on jstor.org. His theorem about the complex roots of a cubic is particularly interesting, although it hangs on a square root of a slope, which seems a bit demanding in a graphical or geometric demonstration.
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