We have a equation: $2x^2+7x+3$
I tried to find the vertex of the parabola by this formula: $a(x-h)^2+k$
but I could not get it.
I got this: but it is not right.
$2(x^2+ \frac{7}{2}x+\frac{49}{4})+3-\frac{49}{2}$
What's the problem? Is it possible to solve it by completing square?
Solve quadratic equation by completing square
quadratics
Best Answer
Completing the square, it is :
$$2x^2 + 7x + 3 = 2\bigg(x^2+ \frac{7}{2}x + \frac{49}{16}\bigg) - \frac{25}{8} = 2\bigg(x + \frac{7}{4}\bigg)^2 - \frac{25}{8}$$
Thus, the vertex can be found at the point where the "completed square term" equals zero :
$$x+\frac{7}{4} = 0 \Leftrightarrow x = -\frac{7}{4}$$
This is truly the vertex point, as it can be cross-validated by the fact that a vertex of a parabola can be found at :
$$x_v = -\frac{b}{2a} \Rightarrow x_v = -\frac{7}{2\cdot 2}$$
Now, the other coordinate of the vertex is the remaining constant term on the right, thus $-25/8$.
This means that the given vertex coordinates, are :
$$(h,k) = \bigg(-\frac{7}{4},-\frac{25}{8}\bigg)$$