Question

Two quadratic functions are shown.

Function 1:
f(x) = 4x2 + 8x + 1
Function 2:
x g(x)
−2 2
−1 0
0 2
1 8


Which function has the least minimum value and what are its coordinates?
Function 1 has the least minimum value and its coordinates are (−1, −3).
Function 1 has the least minimum value and its coordinates are (0, 1).
Function 2 has the least minimum value and its coordinates are (−1, 0).
Function 2 has the least minimum value and its coordinates are (0, 2).

Answer 1

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).

Answer 2

`\bf \qquad\qquad\textit{vertex of a parabola}\\ \quad \\ \begin{array}{llcclll} f(x)=&4x^2&+8x&+1\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)`