Question 760887
<font face="Times New Roman" size="+2">


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


The *[tex \LARGE y]-coordinate of the *[tex \LARGE y]-intercept is given by:


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


Hence the *[tex \LARGE y]-intercept is the point *[tex \LARGE \left(0,c\right)]


The *[tex \LARGE x]-coordinates of the *[tex \LARGE x]-intercepts, if they exist are the real number results of the quadratic formula.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_{1,2}\ =\ \frac{-b\ \pm\sqrt{b^2\ -\ 4ac}}{2a}]


Note:  2 real and distinct roots are guaranteed when the signs on coefficients *[tex \LARGE a] and *[tex \LARGE c] are opposite.


The *[tex \LARGE x]-intercepts are then *[tex \LARGE \left(x_1,0\right)] and *[tex \LARGE \left(x_2,0\right)]


The *[tex \LARGE x]-coordinate of the vertex is *[tex \LARGE x_v\ =\ -\frac{b}{2a}]


The *[tex \LARGE y]-coordinate of the vertex is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v\ =\ f(x_v)\ =\ a(x_v)^2\ +\ b(x_v)\ +\ c]


The axis of symmetry is the vertical line that passes through the vertex.  *[tex \LARGE x\ =\ x_v]


All you have to do is plug in the numbers for your particular *[tex \LARGE a], *[tex \LARGE b], and *[tex \LARGE c] (remembering the binding rules for the minus signs) into the above and do the indicated arithmetic.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>