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March 18th, 2017, 09:27 PM  #1 
Member Joined: Jan 2017 From: US Posts: 65 Thanks: 5  Vertex and Axis Symmetry? f(x)=x$\displaystyle ^2$3x+3 How would you find the vertex and the axis of symmetry? 
March 18th, 2017, 10:06 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,109 Thanks: 580 
You try to get the equation into "vertex" form $f(x) = a(xh)^2 + k$ where the vertex appears at $(h,k)$ the way to do this is known as "completing the square" $x^2  3x + 3 = $ $\left(x  \dfrac 3 2\right)^2 \left(\dfrac 3 2\right)^2 + 3=$ $\left(x  \dfrac 3 2\right)^2 +\dfrac 3 4$ The vertex can be read right off as $v=\left(\dfrac 3 2 , \dfrac 3 4\right)$ For a standard parabola like this, that hasn't been rotated, the axis of symmetry is a vertical line passing through the vertex. This can be described as $x = \dfrac 3 2$ 
March 19th, 2017, 05:04 AM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,429 Thanks: 1196 
for the graph of the parabola $f(x)=ax^2+bx+c$, the vertex is located at coordinates $\bigg(\dfrac{b}{2a},f\left(\dfrac{b}{2a}\right)\bigg)$ $f(x)=x^23x+3 \implies a=1 \text{ and } b=3$ $x= \dfrac{b}{2a}=\dfrac{(3)}{2(1)}=\dfrac{3}{2}$ $y=f\left(\dfrac{3}{2}\right) = \left(\dfrac{3}{2}\right)^2 3\left(\dfrac{3}{2}\right) +3 = \dfrac{9}{4}\dfrac{9}{2}+3 = \dfrac{3}{4}$ axis of symmetry mentioned in romsek's post ... 

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