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Quadratic Functions

Quadratics

Write a program to solve the following problems related to quadratic functions. To do these you are expected to be familiar with quadratics.

For testing, use the equation: y=x22x3

-8-6-4-22468-8-6-4-22468y = x²-2x-3

Notes and Code Outcome

I. Find the Zeros

Find the zeros of a given quadratic. The zeros are the x values when the curve crosses the x-axis (i.e. when y = 0)

Test Case:

-8-6-4-22468-8-6-4-22468y = x²-2x-3(3, 0)(-1, 0)

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II. Coordinates of the Vertex

Find the coordinates (x, y) of the vertex of the quadratic (it's maximum or minimum.)

-8-6-4-22468-8-6-4-22468y = x²-2x-3(1, -4)

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III. Type of Critical Value (Max or Min)

Determine if the critical value is a maximum or minimum.

-8-6-4-22468-8-6-4-22468minimum(1, -4)

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IV. Slope of the line.

Find the slope of a line tangent to the quadratic when: x=2 If you have not taken calculus, see the hint to the right.

-8-6-4-22468-8-6-4-22468

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V. Chord Length

Given two x values (-2 and 2), find the length of a chord between the two points on the curve.

-8-6-4-22468-8-6-4-22468

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VI. Line Length

Estimate the length of the curve between x = -2 and x = 2 by using a series of chords along the line.

-8-6-4-22468-8-6-4-22468

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VII. Line Length: Taking it to the Limit

You can get a better/more accurate estimate of the distance along the curve by breaking the curve into smaller segments. Estimate the length of the curve between x = -2 and x = 2 by using more than just four line segments (use as many as you can).

-8-6-4-22468-8-6-4-22468

We describe the process of adding up the sum of smaller and smaller cords as taking the limit as the change in x (dx) approaches zero; after all, if you made the step zero, you'd have infinite steps.

Mathematically, the limit as dx goes to zero is written as: lim

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Hints

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VIII. Finding the Slope using the chord

You can also use the limit method to estimate the slope of the line.

Say, you want to find the slope when x = 2 (as in section IV). You could just take two points--when x = 1 and x = 2, and find the slope of the chord between those two points.

-8-6-4-22468-8-6-4-22468dx = 1chord slope = 1.0

If we instead take the second point to be when x = 1.5 with the first point still x = 2, the difference in x (dx) is 0.5, and the slope becomes closer to the actual slope:

-8-6-4-22468-8-6-4-22468dx = 0.5chord slope = 1.5

Mathematically, the slope at a point can be found by making dx as small as possible: i.e. taking the limit as dx goes to zero.

Find the slope at x = 2 by finding the slope of a very small chord.

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As we know from section IV, the slope should be 2.

Hints

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