Let's solve this step-by-step.
### Step 1: Determine the Minimum Value of Function 1
Function 1 is given by:
[tex]\[ f(x) = 2x^2 - 8x + 1 \][/tex]
A quadratic function of the form \( ax^2 + bx + c \) has its vertex at:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For \( f(x) = 2x^2 - 8x + 1 \):
- \( a = 2 \)
- \( b = -8 \)
- \( c = 1 \)
The x-coordinate of the vertex is:
[tex]\[ x = -\frac{-8}{2 \cdot 2} = \frac{8}{4} = 2 \][/tex]
The y-coordinate (or the minimum value) of the vertex is found by substituting \( x = 2 \) into \( f(x) \):
[tex]\[ f(2) = 2(2)^2 - 8(2) + 1 = 2(4) - 16 + 1 = 8 - 16 + 1 = -7 \][/tex]
So, the minimum value of Function 1 is \( -7 \) at \( (2, -7) \).
### Step 2: Determine the Minimum Value of Function 2
For Function 2, we are given specific points:
[tex]\[
\begin{array}{|c|c|}
\hline
x & g(x) \\
\hline
-2 & 2 \\
\hline
-1 & -3 \\
\hline
0 & 2 \\
\hline
1 & 17 \\
\hline
\end{array}
\][/tex]
We need to find the minimum \( g(x) \) among these points. Checking the values:
- \( g(-2) = 2 \)
- \( g(-1) = -3 \)
- \( g(0) = 2 \)
- \( g(1) = 17 \)
The minimum value of \( g(x) \) is \( -3 \) at \( (-1, -3) \).
### Step 3: Compare the Minimum Values of Function 1 and Function 2
- The minimum value of Function 1 is \( -7 \) at \( (2, -7) \).
- The minimum value of Function 2 is \( -3 \) at \( (-1, -3) \).
Comparing \( -7 \) and \( -3 \):
[tex]\[ -7 < -3 \][/tex]
Thus, the least minimum value is \( -7 \), and it occurs at coordinates \( (2, -7) \).
### Conclusion
Function 1 has the least minimum value. The least minimum value is [tex]\( -7 \)[/tex] and its coordinates are [tex]\( (2, -7) \)[/tex].
