💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# Find equations of the tangent line and normal line to the curve at the given point.$y^2 = x^3, (1,1)$

Check back soon!

Derivatives

Differentiation

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

It's clear sailing right here. So the curve has two phones functions associated with it, which is plus and minus X to the three house power one has a graph above the X axis, and the other one's below the X axis. We know that why it's bigger than zero at the point. One comma one. So we're gonna take positive X three hands. We have our slope, which is equal to D over D, acts for X 23 house power for access equal to one, and this is equal to three halves and the tangent line equation becomes lie. Minus one is equal to three house X minus one, which is equal to three house acts minus on her. We have our slope of our normal line, which is negative 2/3 since they have to multiply to become negative one. And it's why minus one is equal to 2/3 negative 2/3 X minus one, which is equal to negative 2/3 acts plus 5/3

Derivatives

Differentiation

Lectures

Join Bootcamp