Question

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

93.$y=x^2+2$; at$(1,3)$.

Short answer

The equation of the tangent line is$y=2x+1$.

Step-by-step solution

Step 1. Given information

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

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

$$\begin{gathered} 3=m\left(1\right)+b \\ 3-m=b \\ b=3-m \end{gathered}$$

Substitute $b=3-m$in $y=mx+b$.

$y=mx+(3-m)$.

So the system of equations here is

$$\begin{gathered} y=mx+(3-m) \\ y=x^2+2 \end{gathered}$$.

Step 2. Solve the system of equations obtained.

Substitute $y=mx+(3-m)$in $y=x^2+2$.

$$\begin{gathered} mx+(3-m)=x^2+2 \\ mx+3-m=x^2+2 \\ 0=x^2-mx+(m-1) \\ {x}^2-mx+(m-1)=0 \end{gathered}$$

Using the quadratic formula,

role="math" localid="1646960621570" $$\begin{gathered} x=\frac{mx\pm \sqrt{{\left(m\right)}^2-4\left(1\right)(m-1)}}{2\left(1\right)} \\ x=\frac{mx\pm \sqrt{m^2-4(m-1)}}{2} \end{gathered}$$

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

$$\begin{gathered} m^2{x}^2-4(m-1)=0 \\ {m}^2-4m+4=0 \\ {(m-2)}^2=0m-2=0 \\ m=0 \end{gathered}$$

Step 3. Substitute m=2 in y=mx+b.

We get

$$\begin{gathered} y=2x+(3-2) \\ y=2x+1 \end{gathered}$$

So the equation of the tangent line of$y=x^2+2$at$(1,3)$is$y=2x+1$.