Answer
absolute/local minimum: f(x)=1 and x=2
Work Step by Step
Given $y=2x^2-8x+9$
The critical point is where the first derivative is zero:
$\frac{dy}{dx}=\frac{d{(2x^2-8x+9)}}{dx}$
${\frac{dy}{dx}}=4x-8$
so the critical point is ${\frac{dy}{dx}}=0$
$f'(x)=4x-8=0$
$x=2$
the critical value of y can be obtained by putting in the value of x
$y=2\times2^2-8\times2+9=1$
thus the absolute/local minimum is: f(x)=1 and x=2
