Question Show Examples Find the minimum value of the function $f(x)=2 x^{2}+6 x-1.6$ to the nearest hundredth. Answer Attempt 1 out of 2 Submit Answer
Solution
Step 1 ：The function given is a quadratic function of the form \(f(x)=ax^{2}+bx+c\), where \(a=2\), \(b=6\), and \(c=-1.6\).
Step 2 ：The minimum value of a quadratic function can be found by finding the vertex of the parabola. The x-coordinate of the vertex is given by \(-\frac{b}{2a}\).
Step 3 ：Substituting the values of \(a\) and \(b\) into the formula, we get the x-coordinate of the vertex as \(-\frac{6}{2*2} = -1.5\).
Step 4 ：We can substitute this value into the function to find the y-coordinate, which is the minimum value of the function.
Step 5 ：Substituting \(x = -1.5\) into the function \(f(x)\), we get \(f(-1.5) = 2*(-1.5)^{2} + 6*(-1.5) - 1.6 = -6.10\).
Step 6 ：Thus, the minimum value of the function \(f(x)=2 x^{2}+6 x-1.6\) to the nearest hundredth is \(\boxed{-6.10}\).
