Q. 93 - Precalculus Enhanced with Graphing Utilities

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 $(1,3)$.

Step-by-Step Solution

Answer

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

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,

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

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 = m x + 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 $y=2x+1$.