Question

How do you find the vertex and axis of symmetry for `f (x)= 2x² - 6x+ 3`?

Answer 1

Given and equation of the form: `f(x)=ax^2+bx+c`
The equation for the axis of symmetry is: `x =-b/(2a)`
The x coordinate for the vertex, h, is the same.
The y coordinate, `k=f(h)`

Explanation:

Given: `f(x)=2x^2 - 6x+ 3`

`a = 2, b = -6 and c =3`

The equation for the axis of symmetry is:

`x = -(-6)/(2(2))`

`x = 3/2`

The x coordinate of the vertex is the same:

`h = 3/2`

The y coordinate of the vertex is:

`k = f(h)`

`k = 2(h)^2-6(h)+3`

`k = 2(3/2)^2-6(3/2) + 3`

`k = -3/2`

The vertex is `(3/2,-3/2)`

Answer 2

`(3/2,-3/2),x=3/2`

Explanation:

`"the equation of a parabola in "color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`

`"where " (h,k)" are the coordinates of the vertex"`
`"and a is a constant"`

`"to express f(x) in this form "color(blue)"complete the square"`

`f(x)=2(x^2-3xcolor(red)(+9/4)color(red)(-9/4))+3`

`color(white)(f(x))=2(x-3/2)^2-9/2+3`

`color(white)(f(x))=2(x-3/2)^2-3/2larrcolor(red)" in vertex form"`

`rArrcolor(magenta)"vertex" =(3/2,-3/2)`

`"since " a>0" then minimum "uuu`

`"the axis of symmetry passes through the vertex"` is vertical

`"with equation " x=3/2`
graph{(y-2x^2+6x-3)(y-1000x+1500)=0 [-10, 10, -5, 5]}