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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.
f(x)=x^2+6x-38
if A is an ideal and phi is onto S,then phi(A)is an ideal.
The difference of two numbers is 18. The larger number is four more than twice the smaller. What are the numbers?
Example: We are investing $100,000 in an account that earns interest at a rate of 7.5% for 54 months. Find out how much money will be in the account if, (a) Interest is comp
Example: Solve following equations. 2 log 9 (√x) - log 9 (6x -1) = 0 Solution Along with this equation there are two logarithms only in the equation thus it's easy t
9x^2/-4x^3y^4 x 16x^4y^2/25xy
(-3) EVALUATE X3 -2X2 + 6X -64
how to to a equations ?
how solve the 2, x>-3
-3t^2+15t-18
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