Formula is y = a(x-h)^2 + k
Where h is 1 and k is 1
f (x) = a(x-1)^2 + 1
-3 = a(0-1)^2 + 1
-3 = a(-1)^2 + 1
-3 = a(1) + 1
-3 - 1 = a
-4 = a
a = -4
A must be equal to -4
y = -4(x-1)^2 + 1
0 = -4(x-1)^2 + 1
4(x^2 - 2x + 1) - 1 = 0
4x^2 - 8x + 4 - 1 = 0
4x^2 - 8x + 3 = 0
4x^2 - 8x = -3
Divide fpr 4 each term of the equation....x^2 - 2x = -3/4
We must factor the perfect square ax^2 + bx + c which we don't have. We must follow the rule (b/2)^2 where b is -2....(-2/2)^2 =
(-1)^2 = 1 and we add up that to both sides
x^2 - 2x + 1 = -3/4 + 1
x^2 - 2x + 1 = 1/4
(x-1)^2 = 1/4
square root both sides x-1 = (+/-) 1/2
x1 = +1/2 + 1 = 3/2
x2 = -1/2 + 1 = 1/2
x-intercepts are 1/2 and 3/2, in form (3/2,0); (1/2,0)
Perimeter=106ft
Area=448ft^2
Make sure you remember to subtract various values to find the length of missing sides.
Answer:
Yes
Step-by-step explanation:
There is no unclosable gap. 7+1>7. 7+7>1. An isosceles triangle would be formed.
