Hello,
26r^3s+52r^5-39r²s^4
=13r²(4r^3+2rs-3s^4)
Answer:
We have not write the question properly . so I write the point .
Answer:
see below
Step-by-step explanation:
f(x) = -x^2 +4
The vertex form is
y = a(x-h)^2 +k
Rewriting
f(x) = -(x-0)^2 +4
The vertex is (0,4) and a = -1
Since a is negative we know the parabola opens downward
f(x) = -(x^2 -4)
We can find the zeros
0 = -(x^2 -2^2)
This is the difference of squares
0 = -(x-2)(x+2)
Using the zero product property
x-2 =0 x+2 =0
x=2 x=-2
(2,0) (-2,0) are the zeros of the parabola and 2 other points on the parabola
We have the maximum ( vertex) and the zeros and know that it opens downward, we can graph the parabola
A = 1/2 h (a+b)
A / 1/2 h = a + b
A / 1/2 h - b = a
a = A / 1/2 h - b
Hope this helps
Answer:
both of them thats the answer
Step-by-step explanation: