4th Problem 2016

To find the $x$ value of the vertex, we must use the formula $x = -\frac {b}{2a}$ . We get $x = -\frac {3}{18} = -\frac {1}{6}$ .

To get the $y$ value of the vertex we must plug in the $x$ value of the vertex in the equation. We get $f\left( x \right) =9( -\frac {1}{6} )^{ 2 } + 3(-\frac {1}{6}) +1 = 9(\frac {1}{36}) + 3(-\frac {1}{6}) + 1 = \frac {9}{36} - \frac {3}{6} + 1 = \frac {1}{4} - \frac {1}{2} + 1 = \frac {2}{8} - \frac {4}{8} + \frac {8}{8} = \frac {6}{8} = \frac {3}{4}$ .

Therefore, the vertex of this function is $(-\frac {1}{6} , \frac {3}{4})$ .

This solution is really an extension to one of the solutions below. The vertex of a function is given by setting its derivative to zero and solving for the x value.So for the parabolic case we have ax^2 +bx+c. We can simplify this to x^2+bx/a+c.Taking the derivative here we get 2x+b/a.Setting this equal to zero we get x =-b/2a.Now if substitute 3 and 9 we have the vertex is at -1/6 and since the x values in the answer choices are unique that is all that is required to chose the correct answer.

The vertex of a parabola can be found by turning the parabola into the vertex form. $\begin{aligned} 9x^2 + 3x + 1 & = 9\left(x^2 + \dfrac{1}{3}x + \dfrac{1}{9}\right)\\ & = 9\left(x + \dfrac{1}{6}\right)^2 + 9\left(\dfrac{1}{9} - \frac{1}{36}\right)\\ & = 9\left(x + \dfrac{1}{6}\right)^2 + \dfrac{3}{4} \end{aligned}$

From the vertex form we can see that the vertex is $\boxed{\left(-\dfrac{1}{6}, \ \dfrac{3}{4}\right)}$ .