Problem

Find the minimum value of the function $f(x)=x^{2}+6 x+12$ to the nearest hundredth.

Solution

Step 1 :The function \(f(x)=x^{2}+6 x+12\) is a quadratic function. The graph of a quadratic function is a parabola. If the coefficient of \(x^{2}\) is positive, the parabola opens upwards, and its minimum value is the vertex of the parabola.

Step 2 :The x-coordinate of the vertex of a parabola given by \(f(x)=a x^{2}+b x+c\) is \(-b/2a\). In this case, \(a=1\) and \(b=6\), so the x-coordinate of the vertex is \(-6/2*1=-3\).

Step 3 :Substituting \(x=-3\) into the function will give the minimum value of the function.

Step 4 :The minimum value of the function \(f(x)=x^{2}+6 x+12\) to the nearest hundredth is \(\boxed{3}\).

From Solvely APP
Source: https://solvelyapp.com/problems/7323/

Get free Solvely APP to solve your own problems!

solvely Solvely
Download