Question

find the minimum value of the function $f(x) = x^2 - 4x - 1.2$ to the nearest hundredth.

Explanation:

Step1: Identify the function type

The function \( f(x) = x^2 - 4x - 1.2 \) is a quadratic function in the form \( f(x)=ax^2 + bx + c \), where \( a = 1 \), \( b=-4 \), and \( c=-1.2 \). Since \( a>0 \), the parabola opens upwards, and the vertex is the minimum point.

Step2: Find the x-coordinate of the vertex

The formula for the x-coordinate of the vertex of a quadratic function \( ax^2+bx + c \) is \( x=-\frac{b}{2a} \).
Substituting \( a = 1 \) and \( b=-4 \) into the formula:
\( x = -\frac{-4}{2\times1}=\frac{4}{2} = 2 \)

Step3: Find the y-coordinate (minimum value)

Substitute \( x = 2 \) into the function \( f(x)=x^2 - 4x - 1.2 \):
\( f(2)=(2)^2-4\times(2)-1.2 \)
\( = 4 - 8 - 1.2 \)
\( = -4 - 1.2 \)
\( = -5.2 \)

Answer:

\( -5.20 \)