When f(x) = 0, x = m or n.

So the turning point's x-coordinate must
be the midpoint of m and n, since (0,m) and (0,n) are equidistant from the vertical line of symmetry of the curve, which passes through the turning point

= (m+n)/2

b is a negative constant, so this means the turning point is below the x-axis.


f(x) = p(x-q)² + r
Notice when f(x) = r, p(x-q)² = 0
Then x = q

There is only one value of x when f(x) = r. Since the turning point is the point where there is only one x value for a value of f(x), (any other point on this quadratic curve has 2 values of x for every value of f(x)

The coordinates of the turning point are (q,r)

As the turning point on y = x² corresponds to (a,b) on y = f(x), the turning point on the transformed curve is (a,b)



So from the above we can establish that
q = a, r = b
(m+n)/2 = a

f(x) = p(x-a)² + b
when x = 0, f(x) = c,

c = p(-a)² + b
c = a²p + b
a²p = c-b
p = (c-b)/a²




Since f(x) = p(x-a)² + b, when f(x) = 0,

0 = p(x-a)² + b
-b = p(x-a)²

-b/p = (x-a)²

(x-a) = ± √(-b/p)

(Note that as b is negative, -b is positive. Since c and a are also positive, p is also positive. -b/p is therefore positive and square rooting it gives a real value)

x = a ± √(-b/p)

Sub the values of p,

x = a ± √(-b/((c-b)/a²))

x = a ± √(-a²b/(c-b))

Recall that when f(x) = 0 ,x = m or n.

Since n>m,

n = a + √(-a²b/(c-b)) , m = a - √(-a²b/(c-b))