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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.
I am looking the domain of g^2-6g-55/g. The denominator here can be also be written as 1g, right?
one no. is 7 more than another and its square is 77 more than the square of the smaller number.What are the numbers?
the words and definitions to study please :)
7x+6=x-30
graph the following and find the point of intersection 2x+y=-4 y+2x=3
The "humps" where the graph varies direction from increasing to decreasing or decreasing to increasing is frequently called turning points . If we know that the polynomial con
9x^4-x^2-12x-36
Kevin randomly selected 1 card from a standard deck of 52 cards. what is the probabilty that he will chose King of Hearts ... 17/13 ..... 17/52 .... 13/4 .... 52/12
how do i simplify 1-1/10x divided by 3+7/5x
Actually here we're not going to look at a general cubic polynomial. Here we are jsut going to look at f ( x ) = x 3 . Really there isn't much to do here other than only plugging
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