Question

find the zeros of the function \\( f(x) = 0.1x^2 - 0.4x - 2.2 \\). round values to the nearest thousandth (if necessary).

Explanation:

Step1: Recall Quadratic Formula

For a quadratic function \( f(x) = ax^2 + bx + c \), the zeros are found using \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 0.1 \), \( b=-0.4 \), \( c = -2.2 \).

Step2: Calculate Discriminant

Discriminant \( D = b^2 - 4ac = (-0.4)^2 - 4(0.1)(-2.2) = 0.16 + 0.88 = 1.04 \).

Step3: Apply Quadratic Formula

\( x=\frac{-(-0.4)\pm\sqrt{1.04}}{2(0.1)}=\frac{0.4\pm\sqrt{1.04}}{0.2} \).
Calculate \( \sqrt{1.04}\approx1.0198 \).
First root: \( x_1=\frac{0.4 + 1.0198}{0.2}=\frac{1.4198}{0.2}=7.099 \).
Second root: \( x_2=\frac{0.4 - 1.0198}{0.2}=\frac{-0.6198}{0.2}=-3.099 \).

Answer:

The zeros of the function are \( x \approx -3.099 \) and \( x \approx 7.099 \) (rounded to the nearest thousandth). If we consider one of them (as the interface allows adding a zero, likely expecting both or one, but based on calculation, the two zeros are approximately -3.099 and 7.099).