Question 808869
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In order to find the equation of a parabola, you need at least 3 points on the parabola.  We are given 2 points.  However, because of symmetry, we know that there is a point with a function value of 7 equidistant from the axis of symmetry x = 2 on the horizontal line y = 7.  So we have a third point (4,7) -- since (0,7) is 2 distant from x = 2, we need a point 2 units on the other side, namely (4,7).


Note that we could also have chosen the point (-5,-3) and achieved the same result.


The equation of a general quadratic function is:


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


So if the function value is 7 when x = 0, which is what the point (0,7) means, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(0)^2\ +\ b(0)^2\ +\ c\ = 7]


Hence, *[tex \Large c\ =\ 7]


Also, since the point (-1,-3) is on the graph,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-1)^2\ +\ b(-1)\ +\ c\ =\ -3]


But since we know that *[tex \Large c\ =\ 7], we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ b\ =\ -10]


And since the point (4,7) is on the graph,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(4)^2\ +\ b(4)\ +\ c\ =\ 7]


But since we know that *[tex \Large c\ =\ 7], we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16a\ +\ 4b\ =\ 0]


Now all <b><i>you</i></b> have to do is to solve the 2X2 system:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ b\ =\ -10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16a\ +\ 4b\ =\ 0]


and then once you know the values of a and b to go along with the value of c you already know, you can make the substitutions into


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


and you will have your desired function.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
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