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Now, let's get back to parabolas. There is a basic procedure we can always use to get a pretty good sketch of a parabola. Following it is.
1. Determine the vertex. We'll discuss how to determine this shortly. It's quite simple, but there are several methods for finding it and so will be discussed separately.
2. Find the y-intercept, (0, f (0)) .
3. Solve f ( x ) = 0 to determine the x coordinates of the x-intercepts if they exist.
4. Ensure that you've got at least one point to either side of the vertex. It is to ensure we get a somewhat accurate sketch. If the parabola contains two x-intercepts then already we'll have these points. If it contains 0 or 1 x-intercept we can either just plug in another x value or employ the y-intercept and the axis of symmetry to obtain the second point.
5. Sketch the graph. At this point we've gotten sufficient points to get a quite decent idea of what the parabola will look like.
Factor Theorem For the polynomial P ( x ) , 1. If value of r is a zero of P ( x ) then x - r will be a factor of P ( x ) . 2. If x - r is a factor of P ( x ) then r will
if the roots of a quadratic equation are (-2+sqrt 6) and (-2-sqrt 6), what is the equation in ax^2+bx+c=0 form?
HOW TO PROGRAM QUADRATIC EQUATIONS, INEQUALITIES AND FUNCTIONS INTO A TI 84 PLUS C SILVER EDITION
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I dont really understand it can you help me
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Solve following equations. (a) x 2 -100 = 0 (b) 25 y 2 - 3 = 0 Solution There actually isn't all that much to these problems. To use the square root property a
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Nicole has just finished writing a research paper. She has hired a typist who will type it. The typist charges $3.50 per page if no charts or graphs are used and $8.00 per pag
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