Keywords:
Quadratic equation, vertex shape, parabola
For this case we have to rewrite the given quadratic equation, in the form of vertex, for this, we must take into account that a quadratic equation of the form \(ax ^ 2 + bx + c = 0\), can be rewritten in the form of vertex as: \(y = a (x-h) ^ 2 + k.\) Vertice is the lowest or highest point of the parabola. The vertex is given by: \((h, k)\). So, let: \(f (x) = 4x ^ 2 + 48x + 10\), to find the equation in the form of vertex, we follow the steps below:
Step 1:
We take the common factor to the first two terms of the equation:
\(f (x) = 4 (x^2 + 12x) + 10\)
Step 2:
We work square:
We divide the coefficient of the term \("12x"\) by 2 and its result is squared, that is:
\((\frac {12} {2}) ^ 2 = 36\)
So, we have:
\(f (x) = 4 (x^2 + 12x + 36-36) + 10\)
Step 3:
We simplify:
\(f (x) = 4 (x^2 + 12x + 36) + 10- (4 * 36)\)
Step 4:
We factor:
\(f (x) = 4 (x + 6) ^ 2-134\)
Thus, \(h = -6\ and\ k = -134\)
Answer:
The equation in the form of vertex is: \(f (x) = 4 (x + 6) ^ 2-134\), and the vertex is \((h, k) = (- 6, -134)\)
