Question 1120339
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Post each question separately.  Show the work you have done.  Describe what it is that is causing you difficulty.


I'll help with the first one here.


4.  You are given two points on a graph and asked for the quadratic function.  In order to uniquely determine a quadratic, you need three points.  That is because, in general, to uniquely determine a polynomial function of degree *[tex \Large n] you need *[tex \Large n\ +\ 1] points.


Fortunately, one of the given points is identified as the vertex of the parabola graph.  So, by symmetry, a third point can be determined.  The vertex is at (3,-4) and the other point, (1, 4) is 2 units distant in the *[tex \Large x]-axis.  Hence there is another point on the parabola 2 units on the other side of the vertex with a function value of 4, namely the point (5, 4).


We begin with the general quadratic function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ ax^2\ +\ bx\ +\ c]


Since the point (1, 4) is on the graph, the following must be true about the desired function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(1)\ =\ 4]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(1)^2\ +\ b(1)\ +\ c\ =\ 4]


which simplifies to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ +\ c\ =\ 4]


Similarly, since we know that (3,-4) and (5,4) are on the graph:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ 3b\ +\ c\ =\ -4]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25a\ +\ 5b\ +\ c\ =\ 4]


We now have a 3X3 linear system.  All you need to do is solve the 3X3 system to obtain the necessary coefficients for your function.  Hint:  Use the MDETERM function in Excel (or Numbers if you are on a Mac) to find the values of the four determinants necessary to solve the system using Cramer's Rule.  There are also several online linear system solvers available.


*[illustration QuadFunctionFrom2Points.jpg].
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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