What is the vertex of the function $f(x)=x^2+12 x$ ?
We are given the quadratic function \(f(x) = x^2 + 12x\). This is a parabola opening upwards since the coefficient of \(x^2\) is positive (\(a=1 > 0\)). The vertex of a parabola defined by \(f(x) = ax^2 + bx + c\) is the point \((h, k)\) where the parabola reaches its minimum value. We are given the coefficients \(a = 1\), \(b = 12\), and \(c = 0\). We need to find the coordinates \((h, k)\) of this vertex using standard algebraic techniques.
The x-coordinate of the vertex of a parabola \(f(x) = ax^2 + bx + c\) is given by the formula \(h = -\frac{b}{2a}\). Substituting the known values \(a=1\) and \(b=12\), we calculate:
\[h = -\frac{12}{2 \times 1} = -\frac{12}{2} = -6.\]
Now, to find the y-coordinate \(k\), we evaluate the function at \(h = -6\):
\[k = f(-6) = (-6)^2 + 12(-6) = 36 - 72 = -36.\]
Thus, the vertex is \((-6, -36)\).
We can verify the vertex by completing the square. Start with \(f(x) = x^2 + 12x\). To form a perfect square trinomial, we add and subtract the square of half the coefficient of \(x\), which is \(\left(\frac{12}{2}\right)^2 = 6^2 = 36\):
\[f(x) = (x^2 + 12x + 36) - 36.\]
This simplifies to:
\[f(x) = (x + 6)^2 - 36.\]
Comparing this to the vertex form \(f(x) = a(x - h)^2 + k\), we can identify \(h = -6\) and \(k = -36\), which confirms our previous result. The vertex is \((-6, -36)\).
In physics, the trajectory of a projectile (like a ball thrown in the air) follows a parabolic path. If the height of the ball is described by a quadratic function, finding the vertex tells us the maximum height the ball reaches and at what time it occurs. For instance, if \(h(t) = -t^2 + 6t\) is the height in meters at time \(t\), the vertex identifies the peak of the throw, helping to calculate performance or analyze movement in sports or engineering tasks.