[tex]y = 2x^2 + x + 8[/tex] is a polynomial of even degree, with a positive [tex]x^2[/tex] coefficient, meaning that
- it will have exactly one turning point
- that turning point will be a minimum
So, if the y-coordinate of the turning point is positive, then this polynomial will be positive for all real values of x.
At a turning point, the gradient of y will be equal to 0. The gradient of y is given by
[tex]\frac{\mathrm{d}y}{\mathrm{d}x} = 4x + 1[/tex].
To find the turning point, set this equal to 0 and solve for x:
[tex]4x + 1 = 0 \implies x = -\frac{1}{4}[/tex].
Substituting this value into the equation gives
[tex]y = 2(-\frac{1}{4})^2 + (-\frac{1}{4}) + 8 = \frac{1}{4} - \frac{1}{4} + 8 = 8 \ \textgreater \ 0[/tex].
Since the minimum point of the equation is greater than 0, the equation will be greater than 0 for all real values of x.
