Vikas TU
Last Activity: 7 Years ago
In this illustration we will utilize the bends y=2x2 , and y=x2+1. This is an exceptionally clear case, yet exhibits the strategy for finding the crossing point of two bends well.
Step 1 - since the LHS of both these conditions is the same (y=...) we can liken the two conditions:
2x2=x2+1
This is a genuinely simple condition to illuminate:
Lets make one side equivalent to zero:
- x2 +1=0
Lets move everything crosswise over to the opposite side to dispose of the less signs.
x2 - 1 =0
This is the distinction of two squares, so can be factorized:
(x+1)(x-1)=0
So the x-directions of the crossing point focuses are +1 and - 1.
Step 2 - Now we have to discover the y-facilitates. We do this by connecting the x-qualities to the first conditions. We can utilize it is possible that one, in light of the fact that the lines meet (so they ought to give us a similar outcome!)
At the point when x= +1,
y=2x2
y=2(1)2 =2
At the point when x= - 1
y=2(- 1)2 = 2
So the purposes of crossing point have facilitates (- 1,2) and (1,2) .