Answer:

Step-by-step explanation:
y=x^2-x+1
We want to solve for x.
I'm going to use completing the square.
Subtract 1 on both sides:
y-1=x^2-x
Add (-1/2)^2 on both sides:
y-1+(-1/2)^2=x^2-x+(-1/2)^2
This allows me to write the right hand side as a square.
y-1+1/4=(x-1/2)^2
y-3/4=(x-1/2)^2
Now remember we are solving for x so now we square root both sides:

The problem said the domain was 1/2 to infinity and the range was 3/4 to infinity.
This is only the right side of the parabola because of the domain restriction. We want x-1/2 to be positive.
That is we want:

Add 1/2 on both sides:

The last step is to switch x and y:



Perpendicular: the slope will be -6
Pass the point: Since we know slope is -6, then (-6)*(-3) + ? = 23 -> ? = 5
So answer: y = -6x + 5
In order for two linear lines to be perpendicular, the product of their gradient must be -1.
Let's take y = x + b and y = -x + b
This is already in the form: y = mx + b, where m is 1 and -1 respectively.
Since the product of their gradient is -1, they are said to be perpendicular.
Answer:
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