in such type of questions it's better to know the nature of curve

**General case** :- Let $f(x)=ax^{2}+bx+c$ where $a\neq0$

** a)** concave or convex

1) if $a>0$ then curve will open up (convex nature)

2) if $a<0$ then curve will open down (concave nature)

**b) Nature of roots**

Let $D=b^{2}-4ac$

1) if $D>0$ , then our curve will intersect $x-axis$ at two different points (means two different real roots)

2) if $D=0$ , then our curve will touch $x-axis$ at one point (means two same real roots)

3) if $D<0$ , then our curve won't touch or cut $x-axis$. Means either it will be completely above the $x-axis$(when $a>0$) or it will be entirely below the $x-axis$ (when $a<0$).Hence imaginary roots.

**Original question**

$f(x)=2x^{2}-2x+6$

1) $a=2$ , so $a>0$ means curve will be open up nature

2) $D=-44$ , so $D<0$ means our curve will be entirely above $x-axis$ , mean imaginary roots.

$\frac{\mathrm{d} y}{\mathrm{d} x}=4x-2=0$ , so at $x=\frac{1}{2}$ , we have Max/Min.

although from graph we can directly it would be minima at $x=\frac{1}{2}$ , yet another way is to know the nature of second derivative which is $\frac{\mathrm{d} (4x+2)}{\mathrm{d} x}=4$ , so $+ve$ means minima at $x=\frac{1}{2}$

So at that level from graph we can directly say , either graph will have maxima in $[0,2]$ at $x=0$ or $x=2$ , so just calculate the value at those points and see which one is max.