Given the function:
[tex]f(x)=3x^2-12x+1[/tex]
Let;s find the minimum or maximum value.
To determine if the function has a maximum or minimum, consider the following conditions:
• If the first term of the quadratic function is negative, the function has a maximum.
,
• If the first term of a quadratic function is positive, the function has a minimum.
Here, the first term is positive, therefore, the function has a minimum.
To find the minimum at x, apply the formula:
[tex]x=-\frac{b}{2a}[/tex]
To determine the values of a and b, aply the general quadratic ecpression:
[tex]ax^2+bx+c[/tex]
Now, compare both expressions:
[tex]\begin{gathered} ax^2+bx+c \\ \\ 3x^2-12x+1 \end{gathered}[/tex]
Thus, we have:
a = 3
b = -12
c = 1
Hence, to find the minimum, substitute 3 for a, and -12 for b in the minimum formula above:
[tex]\begin{gathered} x=-\frac{b}{2a} \\ \\ x=-\frac{-12}{2(3)} \\ \\ x=\frac{12}{6} \\ \\ x=2 \end{gathered}[/tex]
Now, to find the minimum, solve for f(2).
Substitute 2 for x in the function and evaluate for f(2):
[tex]\begin{gathered} f(2)=3(2)^2-12(2)+1 \\ \\ f(2)=3(4)-24+1 \\ \\ f(2)=12-24+1 \\ \\ f(2)=-11 \end{gathered}[/tex]
Therefore, the minimum of the function occurs at the point: (2, -11)
When x = 2, y = -11
ANSWER:
Minimum; -11
