Q. 61 - Calculus

Question

Use the definition of the derivative to find the equations of the lines described in Exercises 59-64.The line tangent to the graph of $y=1-x-x^2$ at the point $(1,-1)$

Step-by-Step Solution

Answer

The equation of tangent line is $y=-x$

Step 1.Given information

Given the function $y=1-x-x^2$ and the point $(1,-1)$

Step 2: Calculate the derivative at the point

Calculating, we get

$$\begin{gathered} y=1-x-x^2 \\ y\text{'}=-1-2x \\ y\text{'}(1,-1)=1-2\left(1\right) \\ y\text{'}(1,-1)=-1 \end{gathered}$$

Step 3: Use the point slope form to get the equation

Calculating, we get

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