Question

find the minimum value of the function $f(x) = 1.4x^2 - 7.3x + 3$ to the nearest hundredth. answer attempt 1 out of 2

Explanation:

Step1: Identify the vertex formula for a quadratic function

For a quadratic function \( f(x) = ax^2 + bx + c \), the x - coordinate of the vertex (which gives the minimum or maximum value) is given by \( x = -\frac{b}{2a} \). Here, \( a = 1.4 \) and \( b=-7.3 \).
\( x = -\frac{-7.3}{2\times1.4}=\frac{7.3}{2.8}\approx2.6071 \)

Step2: Substitute the x - value into the function

Now we substitute \( x\approx2.6071 \) into \( f(x)=1.4x^{2}-7.3x + 3 \)
\( f(2.6071)=1.4\times(2.6071)^{2}-7.3\times(2.6071)+3 \)
First, calculate \( (2.6071)^{2}\approx6.797 \)
Then, \( 1.4\times6.797\approx9.516 \)
Next, \( 7.3\times2.6071\approx19.032 \)
So, \( f(2.6071)=9.516-19.032 + 3=-6.516\approx - 6.52 \) (to the nearest hundredth)

Answer:

\(-6.52\)