Find the minimum value of the quadratic function $2x^2 - 8x + 3$.

1. Problem Description

Find the minimum value of the quadratic function $2x^2 - 8x + 3$.

2. Solution Steps

To find the minimum value of the quadratic function $f(x) = 2x^2 - 8x + 3$, we can complete the square.

First, factor out the coefficient of the $x^2$ term from the first two terms:

$f(x) = 2(x^2 - 4x) + 3$.

Now, complete the square inside the parentheses. Take half of the coefficient of the $x$ term, which is $-4$, so half of it is $-2$. Square this value: $(-2)^2 = 4$.

Add and subtract this value inside the parentheses:

$f(x) = 2(x^2 - 4x + 4 - 4) + 3$.

Rewrite the expression inside the parentheses as a squared term:

$f(x) = 2((x - 2)^2 - 4) + 3$.

Distribute the 2:

$f(x) = 2(x - 2)^2 - 8 + 3$.

Simplify:

$f(x) = 2(x - 2)^2 - 5$.

Since the coefficient of the $(x - 2)^2$ term is positive (2), the parabola opens upwards, and the vertex represents the minimum value of the function. The vertex of the parabola is at $x = 2$, and the minimum value of the function is $f(2) = -5$.

Alternatively, we can use the formula for the x-coordinate of the vertex of a parabola given by $f(x) = ax^2 + bx + c$. The x-coordinate of the vertex is $x = -\frac{b}{2a}$. In this case, $a = 2$ and $b = -8$, so

$x = -\frac{-8}{2(2)} = \frac{8}{4} = 2$.

Then, we substitute $x = 2$ into the function to find the minimum value:

$f(2) = 2(2)^2 - 8(2) + 3 = 2(4) - 16 + 3 = 8 - 16 + 3 = -8 + 3 = -5$.

3. Final Answer

The minimum value is -5.