See here : https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1047022.html
Note that when f(x) = y = 1
We have the following
f(-3) = 1 f(-2) = 1 f (-1) = 1 f(1) = 1 f(2) = 1 f(3) = 1
Sum of all c^s = 6
6/7 < 6 ( 1 + x) / (7 + x) < 2
6/7 < 6 (1 + x) / ( 7 + x) divide through by 6
1/7 < (1 + x) / (7 + x)
1 ( 7 +x) < 7 (1 + x)
7 + x < 7 + 7x
0 < 6x
0 < x
And
6 ( 1 + x) / (7 + x) < 2 divide through by 6
(1 + x) /( 7 + x) < 1/3
3 ( 1 + x) < 1 (7 + x)
3 + 3x < 7 + x
2x < 4
x < 2
0 < x < 2
Which is what we assumed
2x^2 + 13x - 4 = 0
x^2 + (13/2) x - 2 = 0 complete the square
x^2 + (13/2)x + 169 / 16 = 2 + 169 / 16
(x + 13/4 )^2 = 201/16 yake both roots
x + 13/4 = ± sqrt (201) / 4
p = - sqrt (201)/4 - 13/4 q = sqrt (201) / 4 - 13 / 4
If I understand this...let an = x
1/9 < 1 /( 7 + x) < 1/7
1/9 < 1 /( 7 + x) 1 / (7 + x) < 1/7
7 + x < 9 7 < 7 + x
x < 2 x > 0
If we want an integer soliution x = an = 1
Vertex = (h , k) = ( -8 , -5)
Axis of symmetry ⇒ x = h ⇒ x = -8
Without proof.....the maximum area given any perimeter is a square with a side = Perimeter / 4 = 82 / 4 = 20.5 m
Correct, FMP !!!!!
D = 25
E = 90 29 F = 65
tan F = DE / EF
tan (65°) = DE / 29
29 * tan (65°) = DE ≈ 62.2
We will have (possible) integer solutions when the discriminant is a perfect square
a^2 - 4*8a = b^2
a^2 - 32a = b^2
a ( a - 32) = b^2
(With a little help from WolframAlpha )
We will have a perfect square when
a roots
-49 x = -7 , 56
-18 x = -6 , 24
-4 x = -4 , 8
0 x = 0
32 x = -16
36 x = -24 , -12
50 x = -40 , -10
81 x = -72 , -9
