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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.
is force applied to the brakes on a car, and stopping distance a direct variation?
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In this section we're going to revisit some of the applications which we saw in the Linear Applications section & see some instance which will require us to solve a quadratic equat
so I''m having trouble. I honestly don''t under stand this. Y=4x+5. y=-1/4x+4 they want me to tell whether the line is parallel, perpendicular or neither I don''t know how.
Tank A contains four times as much as Tank B. Tank C contains 32 L more than Tank A. The three Tanks contain 482 L of water in all. How many liters of water does Tank A contain?
Prove that cosets of an ideal I in a ring R are disjoint or equal
If the rational number x= b/ c is a zero of the n th degree polynomial, P ( x ) = sx n + ...........+ t Where every the coefficients are
f(x)=xsqr-1
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