Regents Question

It is easy to post a screenshot, if you're working with an IBM-based PC and you know how to use programs like MSPAINT. Let me know, if you want instructions on how to do it using Windows features.

Double-click the image below, to expand it.

[attachment=0:15lrmp2l]RegentsExam16.JPG[/attachment:15lrmp2l]
 

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Shame on the University of the State of New York, for implying that equations have roots.

Polynomials have roots.

Functions have zeros.

Equations have solutions.

And, shame on the University of the State of New York, for not providing any scale on their coordinate system. (Do they believe that all grids have squares that are one unit by one unit, or are they just sloppy?)

Question 16 is simply asking for the solutions to the equation where y = 0. In other words, at what values of x does the parabola cross the x-axis?

Here's another way to interpret the equation. y = -x^2 - 2x + 8 is a formula for calculating the y-coordinate for any point on the parabola, when that point's x-coordinate is known.

Plug in the x-coordinate of some point on the parabola, and the polynomial evaluates to the corresponding y-coordinate.

So, they substituted 0 for y, in this formula. So, which points on the parabola have a y-coordinate of zero? Those points are the x-intercepts, right? (Any point on the x-axis has a y-coordinate of zero.)

You can read the equation solutions right off the graph.

Those two values of x are the roots of the quadratic polynomial -x^2 - 2x + 8.

Those two values of x are also the solutions to the equation -x^2 - 2x + 8 = 0.

Those two values of x are also the zeros of the function f(x) = -x^2 - 2x + 8.

Got it? 8-)
 
thank you so much for helping me understand this problem also thank you for putting a screen shot of it.
 
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