Answer:
y=1/2(x-1)
Step-by-step explanation:
If x=t^2 and t>0, then t=sqrt(x).
If t=sqrt(x) or x^(1/2) and y =1-1/t, then y=1-x^(-1/2).
The x-intercept is when y=0.
So we need to solve 0=1-x^(-1/2) to find point P.
Add x^(-1/2) on both sides: x^(-1/2)=1.
Raise both sides to -2 power: x=1
So point P is (1,0).
Let's find tangent line at point (1,0).
We will need the slope so let's differentiate.
y'=0+1/2x^(-3/2)
y'=1/(2x^(3/2))
The slope at x=1 is y'=1/(2[1]^(3/2))=1/(2×1)=1/2.
Recall point-slope form is y-y1=m(x-x1).
So our line we are looking for is y-0=1/2(x-1)
Let's simplify left hand side y=1/2(x-1)
