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Chapter 11: Q. 93 (page 775)

In Problems 92鈥�98, use Descartes鈥檚 method from Problem 91 to find the equation of the line tangent to each graph at the given point.

93.y=x2+2; at(1,3).

Short Answer

Expert verified

The equation of the tangent line isy=2x+1.

Step by step solution

01

Step 1. Given information

The general form of the equation of the tangent must be y=mx+b.

Substitute (x,y)=(1,3)in y=mx+b.

3=m(1)+b3-m=bb=3-m

Substitute b=3-min y=mx+b.

y=mx+(3-m).

So the system of equations here is

y=mx+(3-m)y=x2+2.

02

Step 2. Solve the system of equations obtained.

Substitute y=mx+(3-m)in y=x2+2.

mx+(3-m)=x2+2mx+3-m=x2+20=x2-mx+(m-1)x2-mx+(m-1)=0

Using the quadratic formula,

role="math" localid="1646960621570" x=mx卤(m)2-4(1)(m-1)2(1)x=mx卤m2-4(m-1)2

For the unique solution, the two roots must be equal which means

m2x2-4(m-1)=0m2-4m+4=0(m-2)2=0 m-2=0m=0

03

Step 3. Substitute m=2 in y=mx+b.

We get

y=2x+(3-2)y=2x+1

So the equation of the tangent line ofy=x2+2at(1,3)isy=2x+1.

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