Question

find the minimum value of the function $f(x) = 2x^2 - 11x + 22.4$ to the nearest hundredth.

Explanation:

Step1: Identify the formula for vertex of parabola

For a quadratic function \( f(x) = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x = -\frac{b}{2a} \). Here, \( a = 2 \), \( b=- 11 \), \( c = 22.4 \).
So, \( x=-\frac{-11}{2\times2}=\frac{11}{4} = 2.75 \)

Step2: Substitute x into the function

Substitute \( x = 2.75 \) into \( f(x)=2x^{2}-11x + 22.4 \)
\( f(2.75)=2\times(2.75)^{2}-11\times(2.75)+22.4 \)
First, calculate \( (2.75)^{2}=7.5625 \)
Then, \( 2\times7.5625 = 15.125 \)
\( 11\times2.75 = 30.25 \)
So, \( f(2.75)=15.125-30.25 + 22.4 \)
\( 15.125-30.25=- 15.125 \)
\( -15.125 + 22.4 = 7.275\approx7.28 \) (to the nearest hundredth)

Answer:

\( 7.28 \)