Help with num 4 please.​

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)

00

step-by-step explanation:

answer: 0.1,0.85

step-by-step explanation: