Question 611676
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In order to find the *[tex \LARGE x]-intercepts, you need to write the specific function that describes a parabola that has the specific criteria given.  In order to create a unique function whose graph is a parabola you need three points.  The bad news is, you were only given two points, the vertex and the *[tex \LARGE y]-intercept.  But the good news is, one of the points given is the vertex and parabolas are symmetric about a line passing through the vertex.  In this case I'm going to assume that the axis of symmetry is a vertical line.  If that is not the case, then 1) this problem is at least one order of magnitude more difficult than the one I'm going to solve, and 2) the problem cannot be solved with only the information given.


Assuming a vertical line of symmetry, namely *[tex \LARGE x\ =\ 3], there must be a point on the graph with a function value equal to the *[tex \LARGE y]-coordinate of the *[tex \LARGE y]-intercept at a horizontal distance from the vertex equal to the horizontal distance of the vertex from the *[tex \LARGE y]-axis.


The *[tex \LARGE y]-axis is 3 units horizontally distant from the vertex at (3,-2).  Therefore there must be another point on the graph of the parabola 3 units on the OTHER side of the vertex, namely at (6,7).


From this information we can create the quadratic function which describes the set of ordered pairs that is the graph of the desired parabola.


The standard form of a quadratic function is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ ax^2\ +\ bx +\ c]


substituting the values of the coordinates of any point in *[tex \LARGE \mathbb{R}^2] for *[tex \LARGE x] and *[tex \LARGE y] in the function results in a true statement if and only if the point selected is on the graph of the function.


Hence, if the point (0,7) is on the graph then


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


which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ 7]


must be a true statement.


Likewise, and I'll leave verification of the arithmetic to you:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ 3b\ +\ c\ =\ -2]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 36a\ +\ 6b\ +\ c\ =\ 7]


must also be true statements.


Using the reduced form of the first result of this analysis we can reduce the 3X3 system of equations to a 2X2 system:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 36a\ +\ 6b\ =\ 0]


And I'll leave it as an exercise for you to determine that the solution to the original 3X3 system is the ordered triple *[tex \LARGE (1, -6, 7)]


resulting in the quadratic function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^2\ -\ 6x\ +\ 7]


From here, set the function equal to 0 (the *[tex \LARGE x]-intercepts are the two places where the value of the function is 0), and solve using ordinary means.  Don't spend much time trying to factor this since it does not factor over the rational numbers.  I recommend the quadratic formula.


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