Question 187108
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First of all, *[tex \Large x + 6] and *[tex \Large x + 4] are <i>not</i> the roots of 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y = x^2 + 10x + 24]


They are the <i><b>factors</b></i> of the trinomial.  <i>x</i> = -6 and <i>x</i> = -4 are the roots of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 + 10x + 24 = 0]


The axis of symmetry of any polynomial equation of the form:


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


is the line *[tex \Large x = \alpha] where *[tex \Large \alpha] is the <i>x</i> coordinate of the vertex.  The <i>x</i> coordinate of the vertex is given by either:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{-b}{2a}]


or the average of the roots:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{x_1 + x_2}{2}]


Where *[tex \Large x_1] and *[tex \Large x_2] are the roots of



*[tex \LARGE \ \ \ \ \ \ \ \ \ \  ax^2 + bx + c = 0]


In your case:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \alpha = \frac{-b}{2a} = \frac{-10}{2} = -5]


or 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \alpha = \frac{-6 + (-4)}{2} = -5]


and the equation for the axis of symmetry is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = -5]


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