Question 274307
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If the equation of the axis is *[tex \Large x\ -\ 3\ = 0] or alternatively *[tex \Large x\ =\ 3], then the *[tex \Large x]-coordinate of the vertex must be 3.


In the equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ a(x\,-\,h)^2\ +\ k]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (h, k)] is the vertex.


Therefore, the value of *[tex \Large h] is given directly and we now have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ a(x\,-\,3)^2\ +\ k]


If the *[tex \Large y]-intercept is 10, that means the graph includes the point (0,10).  In other words:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\ =\ a(0\,-\,3)^2\ +\ k]


which can be written:


<b>1:</b> *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ +\ k\ =\ 10]


if one of the *[tex \Large x]-intercepts is 2, then the graph includes the point (2,0).  In other words:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0\ =\ a(2\,-\,3)^2\ +\ k]


which can be written:


<b>2:</b> *[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ k\ =\ 0]


Multiply Equation 2 by -1:


<b>2a:</b> *[tex \LARGE \ \ \ \ \ \ \ \ \ \ -a\ -\ k\ =\ 0]


Add Equation 2a to Equation 1:


<b>1a:</b> *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9a\ -\ a\ +\ k\ -\ k\ =\ 10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8a\ =\ 10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ \frac{10}{8}]


Then from Equation 2 we see that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ -\frac{10}{8}]


And finally the desired equation is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{10}{8}(x\,-\,3)^2\ -\ \frac{10}{8}]


Check:


Since *[tex \Large (h, k)] is the vertex, *[tex \Large k] is the *[tex \Large y]-coordinate of the vertex.  Since the *[tex \Large x]-coordinate of the vertex was determined to be 3, the value of the function, which is to say the value of *[tex \Large y] must be *[tex \Large -\frac{10}{8}] whenever *[tex \Large x] is 3:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{10}{8}(3\,-\,3)^2\ -\ \frac{10}{8}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 0 -\ \frac{10}{8}]:  Checks


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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