Question

Minimum value of the quadratic equation x^2-3x+5=0 is?

Answer 1

`11/4`

Explanation:

`"to find the minimum value we require to find the vertex"`
`"and determine if max/min"`

`"for a quadratic in "color(blue)"standard form";ax^2+bx+c`

`"the x-coordinate of the vertex is "`

`x_(color(red)"vertex")=-b/(2a)`

`x^2-3x+5" is in standard form"`

`"with "a=1,b=-3" and "c=5`

`x_("vertex")=-(-3)/2=3/2`

`"substitute this value into the equation for y-coordinate"`

`y_("vertex")=(3/2)^2-3(3/2)+5=11/4`

`color(magenta)"vertex "=(3/2,11/4)`

`"to determine whether max/min"`

`• " if "a>0" then minimum "uuu`

`• " if"a<0" then maximum "nnn`

`"here "a=1>0" hence minimum"`

`"minimum value of "x^2-3x+5" is "11/4`
graph{x^2-3x+5 [-10, 10, -5, 5]}