## University Calculus: Early Transcendentals (3rd Edition)

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