I think 140 is the answer
To find the vertex of a quadratic function, you need to 'complete the square';
Completing the square gives a quadratic in another form;
For quadratics that have an x² coefficient of 1 as is the case, it is a simple matter to complete the square:
For the quadratic x² + bx + c
Simply, put x added to half the coefficient of the x term (b/2) in a bracket and square the bracket;
Then add the constant (c) and the square of the half the coefficient of the x term (b/2), so:
(x + d)² + e, where d is a (rational) number to be identified;
d = b/2
e = d² + c
So, for f(x) = x² + 0x + 3:
Then the completed-the-square form is: f(x) = (x + 0)² + 3
As it happens, this quadratic function has a common normal form and completed-the-square form because the x-coefficient is 0.
The vertex coordinates are, according to the general format given above:
(-d, e).
So, looking at completed-the-square form for the function in question, we can tell the vertex is at the coordinates: (0, 3)
Quadratic functions always have symmetry about the vertical line with the x-coordinate of the vertex;
This makes sense if you think about how the graph looks (u-shaped);
Therefore for our function, the vertex is:
x = 0 (a vertical line with the x-coordinate of the vertex).
It should be 17, 26, 35, 44, 53
