Vertexed

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 Ashish's solution
2. Complete the square and write in the form $f(x) = a(x-p)^2 + q$ . The vertex is given as $(p,q)$ , as shown in Alex's solution
3. Use Calculus. At vertexes, $f'(x) = 0$

For this question, I will demonstrate Method 3

$f(x) = -2x^2+5x-5\\ f'(x) = -4x+5\\ f''(x) = -4 < 0$

The parabola has a maximum point (a type of vertex) when $f'(x) = 0$ and $f''(x) < 0$ (this has been shown above)

$f'(x) = 0\\ -4x+5=0\\ x=\dfrac{5}{4}$

$f\left(\dfrac{5}{4}\right)\\ = -2\left(\dfrac{5}{4}\right)^2 +5\left(\dfrac{5}{4}\right)-5\\ = -\dfrac{25}{8}+\dfrac{25}{4}-5\\ =-\dfrac{15}{8}$

Therefore, the vertex is $\boxed{\left(\dfrac{5}{4},-\dfrac{15}{8}\right)}$

For a vertex at point $(h,k)$ you must first put the parabola in the form $\large f(x)=a(x-h)^2+k$ We can achieve this by completing the square. $\large y=-2x^2+5x-5$ $\large \frac{y}{-2}=x^2-\frac{5x}{2}+\frac{5}{2}$ $\large \frac{y}{-2} -\frac{5}{2}=x^2-\frac{5x}{2}$ $\large \frac{y}{-2}-\frac{5}{2}+\frac{25}{16}=x^2-\frac{5x}{2}+\frac{5^2}{4^2}=(x-\frac{5}{4})^2$ $\large \frac{y}{-2}=(x-\frac{5}{4})^2+\frac{15}{16}$ and finally $\large f(x)=-2\left(x-\frac{5}{4}\right)^2-\frac{15}{8}$ Notice that the vertex is at the maximum when $x-h=0$ .

The graph's minimum value is $\dfrac{4ac- b^2}{4a} = \dfrac{4×-2×-5 - 5^2}{4×-2} = -\dfrac{15}{8}$ .
And this value is at $-\dfrac{b}{2a} = -\dfrac{5}{2×-2} = \dfrac{5}{4}$ .

So, this graph gived minimum value at $\color{#3D99F6}{\boxed{\left(\dfrac{5}{4},-\dfrac{15}{8}\right)}}$ .