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March 18th, 2017, 10:27 PM   #1
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Vertex and Axis Symmetry?

f(x)=x$\displaystyle ^2$-3x+3

How would you find the vertex and the axis of symmetry?
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March 18th, 2017, 11:06 PM   #2
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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

$x = \dfrac 3 2$
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March 19th, 2017, 06:04 AM   #3
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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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