My Math Forum Vertex and Axis Symmetry?

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 March 18th, 2017, 09:27 PM #1 Senior Member   Joined: Jan 2017 From: US Posts: 112 Thanks: 6 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: USA Posts: 2,040 Thanks: 1063 You try to get the equation into "vertex" form $f(x) = a(x-h)^2 + k$ 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 $x = \dfrac 3 2$ Thanks from Indigo28
 March 19th, 2017, 05:04 AM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,761 Thanks: 1416 for the graph of the parabola $f(x)=ax^2+bx+c$, the vertex is located at coordinates $\bigg(\dfrac{-b}{2a},f\left(\dfrac{-b}{2a}\right)\bigg)$ $f(x)=x^2-3x+3 \implies a=1 \text{ and } b=-3$ $x= \dfrac{-b}{2a}=\dfrac{-(-3)}{2(1)}=\dfrac{3}{2}$ $y=f\left(\dfrac{3}{2}\right) = \left(\dfrac{3}{2}\right)^2 -3\left(\dfrac{3}{2}\right) +3 = \dfrac{9}{4}-\dfrac{9}{2}+3 = \dfrac{3}{4}$ axis of symmetry mentioned in romsek's post ...

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