87th Problem 2016

We can write the vertex of a quadratic function in the form of $f(x) = ax^2 + bx + c$ in an ordered pair of the form $(-\dfrac{b}{2a}, f(-\dfrac{b}{2a}))$ .

$\frac{4}{4}=1\implies(1, f(1))\implies(1,-8) \rightarrow 1 - 8 = \boxed{-7}$ .

There are $3$ methods to find the vertex of quadratic functions $f(x) = ax^2 + bx + c$

1. The coordinates of the vertex is defined as $\left(-\dfrac{b}{2a},f\left(-\dfrac{b}{2a}\right)\right)$ , as shown in the solution by Ralph
2. Complete the square and write in the form $f(x) = a(x-p)^2 + q$ . The vertex is given as $(p,q)$
3. Use Calculus. At vertexes, $f'(x) = 0$

For this question, I will demonstrate Method 2

$f(x) = 2x^2 - 4x - 6\\ =2(x^2 - 2x) -6\\ =2(x^2 - 2x + \left(\dfrac{-2}{2}\right)^2 - \left(\dfrac{-2}{2}\right)^2 ) -6\\ =2\left((x-1)^2 - 1\right) - 6\\ =2(x-1)^2 -2-6\\ =2(x-1)^2 - 8$

The vertex is $(1,-8)$

$x=1,\;y=-8,\;x+y=1+(-8) = \boxed{-7}$