Question

What is the minimum of: y = 2x^2 + 4x - 1

Answer 1

To find the minimum of the quadratic function \( y = 2x^2 + 4x - 1 \), we can use the vertex formula, since a quadratic function in the form \( y = ax^2 + bx + c \) has its vertex at \( x = -\frac{b}{2a} \).

Here, we have:

- \( a = 2 \)
- \( b = 4 \)
- \( c = -1 \)

First, we calculate the \( x \)-coordinate of the vertex:

\[
x = -\frac{b}{2a} = -\frac{4}{2 \cdot 2} = -\frac{4}{4} = -1
\]

Next, we can find the corresponding \( y \)-coordinate by substituting \( x = -1 \) back into the equation:

\[
y = 2(-1)^2 + 4(-1) - 1
\]
\[
= 2(1) - 4 - 1
\]
\[
= 2 - 4 - 1
\]
\[
= -3
\]

Thus, the minimum value of the function \( y = 2x^2 + 4x - 1 \) occurs at \( x = -1 \) and the minimum value is \( -3 \).

Therefore, the minimum of the function is:

\[
\boxed{-3}
\]