Question

find the minimum value of the function $f(x) = 2x^2 - 4.5x - 5$ to the nearest hundredth.

Explanation:

Step1: Identify the vertex formula

For a quadratic function \( f(x) = ax^2 + bx + c \), the x - coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a = 2 \), \( b=- 4.5 \), \( c = - 5 \).

Step2: Calculate the x - coordinate of the vertex

Substitute \( a = 2 \) and \( b=-4.5 \) into the formula \( x=-\frac{b}{2a} \). We get \( x =-\frac{-4.5}{2\times2}=\frac{4.5}{4} = 1.125 \).

Step3: Find the minimum value (y - coordinate)

Substitute \( x = 1.125 \) into the function \( f(x)=2x^{2}-4.5x - 5 \).
\( f(1.125)=2\times(1.125)^{2}-4.5\times1.125-5 \)
First, calculate \( (1.125)^{2}=1.265625 \), then \( 2\times1.265625 = 2.53125 \)
Next, \( 4.5\times1.125 = 5.0625 \)
Then \( f(1.125)=2.53125-5.0625 - 5=2.53125-(5.0625 + 5)=2.53125 - 10.0625=-7.53125\approx - 7.53 \) (to the nearest hundredth)

Answer:

\(-7.53\)