Set the equations equal to find where they intersect: ( I changed 'b' to 'n' for clarity)
x^2 + 2x+7 = 7x+n
x^2 -5x + (7- n) = 0 for there to be only ONE point of intersection, the discrimnant = 0
b^2 - 4ac = 0
-5^2 - 4 (1)(7-n) = 0
25 -28 + 4n = 0
n = 3/4
Here is a graph:
https://www.desmos.com/calculator/boktuzssdd
Remember cos2 + sin2=1 so sin Z = 24/25
Now you can draw a right triangle with sides 7 24 and hypotenuse 25
The Tan y = opp/adj = 7/24
As x starts at some infintesimal value greater than 0 , the funtion is + inf
as x goes to infinity the funtion approaches 0
(0, + inf) would be the range of function values
Sorry.... some of those results will be irrational (I was thinking 'imaginary')....so ...
121 - 4c = rational
30,24,18,10 I think are the ones that will result in a perfect square discriminant
example 121 - 4(18) = 49 a perfect square
In any triangle, all of the angles sum to 180o
soooo....
A + B + C = 180
x + 2x +8x = 180
11x = 180
x = 180/11 =..........
Area = 72 unit2 I think
Type it into the web2 calculator
hit =
look at the dark blue area above the calculator display
5/7 as "guest'' found ....
Real and rational maeans the discriminant of this quadrtatic must be > 0
b^2 - 4ac >0
112 - 4 (1)(c) >0
121>4c
4c<121
c< 30.25 I 'll let you make the list 30,29,........1
-b+3=b add 'b' to both sides of the equation (to keep things 'balanced')
3 = 2b now divide both sides by 2
3/2 = b or b = 3/2
This is a bowl shaped parabola....minimum will be at the vertex
the t value at the vertex is given by - b/2a = - -8/(2(1)) = 4 = t
