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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.
2 3/4 + 1 1/4 =
w^2 + 30w + 81= (-9x^3 + 3x^2 - 15x)/(-3x) (14y = 8y^2 + y^3 + 12)/(6 + y) ac + xc + aw^2 + xw^2 10a^2- 27ab + 5b^2 For the last problem I have to incorporate the following words
i need help on my homework could u help me ?
hello! at my school we are learning how to solve two-step equations but i am having a little trouble. can you please help me?
(2x^2+16x+24)/(3x^2) /(x+6)
Sam is twice as old as John was two years ago. If the difference between their ages be 2 years, how old is Sam today?
Can you get me more questions to practice on this.
The point in graph to the right are (1998),177),(20087),(2002,195 and (2004,207)where the y coordinates are the thousands complete part(a)and (b)(a use the first and last data poin
b^4* b^6
x plus 6 equal -10
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