find the minimum point of f(x)=2x^2 + 8x + 1 (type an ordered pair. simplify your answer.)

Explanation:

Step1: Find the derivative

The derivative of $f(x)=2x^{2}+8x + 1$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=4x + 8$.

Step2: Set the derivative equal to zero

Set $f'(x)=0$, so $4x+8 = 0$. Solving for $x$: Subtract 8 from both sides: $4x=-8$. Then divide by 4, we get $x=-2$.

Step3: Find the y - value

Substitute $x = - 2$ into the original function $f(x)=2x^{2}+8x + 1$. $f(-2)=2(-2)^{2}+8(-2)+1=2\times4-16 + 1=8-16 + 1=-7$.

Answer:

$(-2,-7)$