Question 330684
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Since there are two *[tex \Large x] intercepts, there are two distinct real number roots, hence the discriminant is positive.


However, there is insufficient information to determine the specific quadratic function or its zeros.


Using the coordinates of the vertex, you can determine two vital pieces of information. First since the *[tex \Large x]-coordinate of the vertex of the general quadratic function is given by *[tex \Large \frac{-b}{2a}], you can use the fact that the *[tex \Large x]-coordinate of the vertex is *[tex \Large 0.16] to write a linear equation relating *[tex \Large a] and *[tex \Large b]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.32a\ -\ b\ =\ 0]


and you can also write another linear equation relating *[tex \Large a], *[tex \Large b], and *[tex \Large c] by using the vertex coordinates since we know that any point on the function *[tex \Large f(x)] can be described as *[tex \Large \left(x,\,f(x)\right)]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-0.16)^2\ +\ b(-0.16) +\ c\ =\ -2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.256a\ -\ 0.16b\ +\ c\ =\ -2]


and by substitution:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -0.256a\ +\ c\ =\ -2]


But that is as far as we can go without additional information.  The only thing that can be said for certain about the zeros of the function is that there exists a real number *[tex \Large \alpha] such that the *[tex \Large x]-intercepts of the graph of your function are at:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-0.16\,+\,\alpha,\,0\right)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-0.16\,-\,\alpha,\,0\right)]



The coordinates of any other point on the graph of your function would be sufficient to uniquely determine the coefficients.


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