Question

How do you find the vertex of the parabola `y=2x^2-8x+7`?

Answer 1

You can find the coordinates of your vertex either using a mnemonic or the derivative:

1) Mnemonic: the coordinates of the vertex of a parabola in the form:
`y=ax^2+bx+c` are:
`x_v=-b/(2a)`
`y_v=-Delta/(4a)` where `Delta=b^2-4ac`
in your case:
`x_v=2`
`y_v=-1`

2) Derivative: determine the derivative of your function and set it equal to zero:
`y'=4x-8`
`4x-8=0` gived `x=x_v=2`
substituting back in your function:
`y=y_v=2*4-8*2+7=-1`

Graphically:
graph{2x^2-8x+7 [-4.385, 4.386, -2.19, 2.193]}

Answer 2

A slight variant on method

`color(blue)("Vertex" ->(x,y)->(2,-1)`

Explanation:

Given:`" "2x^2-8x+7`

Write as:`" " 2(x^2-color(red)(8/2)x)+7`

Consider the `color(red)(-8/2)" from "-8/2x`

`color(blue)(x_("vertex")=(-1/2)xx(color(red)(-8/2)) = +8/4 = 2)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute `x=color(magenta)(2)` in the original equation to find `y_("vertex")`

`y=2x^2-8x+7" "->" "y_("vertex")=2(color(magenta)(2))^2-8(color(magenta)(2))+7`

`color(blue)(y_("vertex")= 8-16+7=-1)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Vertex" ->(x,y)->(2,-1)`