Question 273706
<font face="Garamond" size="+2">


I'm not sure what you are trying to say as to your textbook instructions.


However, here is how to find the vertex of a parabola when you have a quadratic function in standard form:


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


The *[tex \Large x]-coordinate of the vertex is given by:


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


and then the *[tex \Large y]-coordinate of the vertex is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v = f(x_v) = f\left(\frac{-b}{2a}\right)]


So for your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_v = \frac{-6}{2(-3)}\ =\ 1]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v = f(1) = -3(1)^2\ +\ 6(1)\ -\ 1]


You can do your own arithmetic.


And look at a graph of your function to check your work:


{{{drawing(
500, 500, -5, 5, -5, 5,
grid(1),
graph(
500, 500, -5, 5, -5, 5,
-3x^2+6x-1))}}}


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