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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.
Consider the function y = 2x. the domain is restricted to 0 = x = 4, what is the range of this function
using linear algebra calculate the equilibrium P 1 ,P 2 ,P 3 for the following three good market model. (1) Qs1=-7+P1 (2) Qd1=15-P1+2P2+P3 (3) Qs1=Qd1 (4) Qs2=-4+4P2
i need with algebra
We have to note a couple of things here regarding function composition. Primary it is NOT multiplication. Regardless of what the notation may recommend to you it is simply not
5(8-2t)
As a last topic in this section we have to briefly talk about how to take a parabola in the general form & convert it into the following form
Assume that P ( x ) is a polynomial along with degree n. Thus we know that the polynomial have to look like, P ( x ) =ax n
3radical 8x^4y^5
How do i do an fx problem?
(-5,-2) y=5/2x+3
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