Question

Find range $y=2x^2-3x+1$ when $xϵ[-3,0]$

Answer 1

Given,

$y=2x^2-3x+1$

Consider,

$2x^2-3x+1=0$

$2x^2-2x-x+1=0$

$2x(x-1)-1(x-1)=0$

$(x-1)(2x-1)=0$

$x=1,\frac{1}{2}$

$\therefore y=[1,\frac{1}{2}]$