Final answer:
Function 1 has the least minimum value with coordinates (-1, -3). The vertex of its quadratic function was found using the vertex formula and substituting the x-coordinate back into the function. Function 2's minimum value is higher, at coordinates (-1, 0), as inferred from the given table of values.
Step-by-step explanation:
To determine which function has the least minimum value, we must first find the minimum value for each function. For Function 1, f(x) = 4x² + 8x + 1, we can find the vertex of the parabola by completing the square or using the formula x = -b/(2a), which gives us the x-coordinate of the vertex.
In this case, a = 4 and b = 8, so the x-coordinate of the vertex is x = -8/(2 × 4) = -1. Substituting this back into the function, we find the minimum value: f(-1) = 4(-1)2 + 8(-1) + 1 = 4 - 8 + 1 = -3. Therefore, the coordinates of the minimum for Function 1 are (-1, -3).
Function 2 does not provide an equation, but a table of values. We notice that as the x-values increase, the g(x) values go from 2 down to 0 and then back up, which suggests that x = -1 could be the vertex, giving the coordinates of the minimum for Function 2 as (-1, 0).
Comparing the minimum values, Function 1's minimum is -3 and Function 2's is 0, so Function 1 has the least minimum value, and the coordinates of its minimum are (-1, -3).