To find the minimum value of the quadratic function $f(x) = 2x^2 + bx + 7$, where $b = -4$, we follow these steps:

1. Determine the vertex formula:
   For a quadratic function $f(x) = ax^2 + bx + c$, the $x$-coordinate of the vertex is given by:
   $x = -\frac{b}{2a}$

2. Substitute $a = 2$ and $b = -4$:
   $x = -\frac{-4}{2(2)} = \frac{4}{4} = 1$

3. Find $f(x)$ at $x = 1$:
   Substitute $x = 1$ into $f(x) = 2x^2 - 4x + 7$:
   $f(1) = 2(1)^2 - 4(1) + 7$ $f(1) = 2 - 4 + 7 = 5$

4. Conclusion:
   The minimum value of the function is 5.00 (rounded to two decimal places).

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Related Questions:

1. What is the effect of changing the coefficient $b$ on the vertex of a parabola?
2. How do you determine whether a quadratic function opens upwards or downwards?
3. Can you find the minimum value of $f(x) = 3x^2 - 6x + 8$?
4. What happens to the minimum value if $a$ becomes negative?
5. How do you calculate the $y$-intercept of the function $f(x) = ax^2 + bx + c$?

Tip:

The sign of $a$ determines whether the quadratic function has a minimum ($a > 0$) or a maximum ($a < 0$).