We are asked to find the minimum value of the quadratic function $2x^2 - 8x + 3$.

1. Problem Description

We are asked to 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 or use the vertex formula. Let's complete the square.

First, factor out the coefficient of $x^2$ 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 $-4/2 = -2$, and square it, $(-2)^2 = 4$. Add and subtract this value inside the parentheses:

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

Rewrite the first three terms inside the parentheses as a perfect square:

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

Distribute the 2:

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

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

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

The minimum value of the function is $-5$.

Alternatively, we can use the vertex formula. For a quadratic function of the form $ax^2 + bx + c$, the x-coordinate of the vertex is given by $x = -\frac{b}{2a}$. In this case, $a = 2$, $b = -8$, and $c = 3$. So,

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

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

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

Thus, the minimum value of the quadratic function is $-5$.

3. Final Answer

-5