The "standard" parabola with roots 0 and 2 is
All multiples of this parabola, i.e.
have the same roots. We can choose the factor such that the parabola passes through the desided point: if we plug 1, 5 for x, y we have
So, our claim is that the parabola
has roots 0 and 2 and vertex at (1, 5).
You can easily verify this: the roots are guaranteed by the fact that we can write the equation as
The vertex must be at x=1, because it's the midpoint of the roots. Moreover, if we evaluate the function at x=1 we have
as required.
Answer:
B
Step-by-step explanation:
Given
(x - 5)²
= (x - 5)(x - 5)
Each term in the second factor is multiplied by each term in the first factor, that is
x(x - 5) - 5(x - 5) ← distribute both parenthesis
= x² - 5x - 5x + 25 ← collect like terms
= x² - 10x + 25 → B
find "r", then plug it in the circle's equation
Answer:
<E=<G=115°
since opposite angle of rhombus is equal