Question
Find the equation of the normal to the curve $x+y=x^{y}$, where it cuts the $x$-axis.
Step 1
This happens when $y=0$. Substituting $y=0$ into the equation of the curve $x+y=x^{y}$, we get $x+0=x^{0}$, which simplifies to $x=1$. So, the point where the curve cuts the x-axis is $(1,0)$. Show more…
Show all steps
Your feedback will help us improve your experience
Varsha Aggarwal and 68 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Find the equation of the normal to the curve $y=\mid x^{2}-\lfloor x \|$ at $x=-2$.
The Tangent and Normal
Level I
Find the equation of the normal to the curve $y=x^{2}-x-2$ at the point $(1,-2)$
Find the equation of the normal to the curve $y=x^{2}+4 x-2$ at the point where $x=-3 .$ Find the coordinates of the other point where this normal intersects the curve again.
Differential Calculus I: Fundamentals
tangents and normals
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD