Step-by-step explanation:
x-intercepts mean points of the curve, where y = 0.
since we have 2 x-intercepts, we need 2 possibilities to bring the functional value to 0.
remember, when is an expression of multiplications 0 ?
when at least one factor is 0, because 0 × something is always 0.
and so we use a little trick :
we say that the quadratic equation is simply the product of 2 terms that are 0 for 1 of the 2 x-intercepts.
for x = 4 we get (x-4).
as you can easily see, this is 0 for x = 4.
for x = -1 we get (x+1)
our basic quadratic equation that has the requested x- intercepts is therefore
y = (x-4)(x+1)
but still, there are infinitely many quadratic equations with the same x-intercepts :
y = a(x-4)(x+1)
we need to pick the one that goes through the given extra point (1, -2) :
-2 = a(1-4)(1+1) = a×-3×2 = a×-6
2 = 6a
a = 2/6 = 1/3
and we get as our final equation :
y = 1/3 × (x - 4)(x + 1) =
= 1/3 × (x² - 3x - 4) = x²/3 - x - 4/3
