The second partial derivatives of exist everywhere in its domain. By Schwarz's theorem, the mixed second-order partials derivatives are equal:
Then
We don't actually need to compute and
to know that they are both free of
. So when we differentiate for a second time with respect to
, the whole thing cancels out.
Answer:
This question is clearly incomplete, so i will answer it in a really general way.
Suppose that for a quadratic function, we know that the x-intercepts are a and b.
And we also know that this function passes through the point (c, d).
First a definition, for a n-degree polynomial with the x-intercepts {x₁, x₂, ...,xₙ} and a leading coefficient K, we can write this polynomial in the factorized form as:
p(x) = K*(x - x₁)*(x - x₂)*...*(x - xₙ)
Now let's do the same for our quadratic function, we can write it as:
f(x) = K*(x - a)*(x - b)
(where a and b are known numbers)
Now we also know that this function passes through the point (c, d)
This means that:
f(c) = d
then:
d = K*(c - a)*(c - b)
With this equation we can find the value of K,
K = d/( (c-a)*(c - b))
Then the quadratic function is:
Where again, it is supposed that you know the values of a and b, and also the point (c, d)