<p>Find the tangent and normal lines to y=x^3 at the point (1,1).</p>
<p>I think I understand this, but I don't want to keep doing it wrong if I am.</p>
<p>I got:
Tangent: y=3x-2
Normal: y=(-1/3)x+4/3</p>
<p>Is that correct? If not what should I do?</p>
<p>those seem right.</p>
<p>yes, you are correct:</p>
<p>derivative of x^3 is 3x^2
plug your x=1 into derivative to get slope: 3(1)^2=3
y=mx+b
plug in your point (1,1)
1=(3)(1)+b, b=-2
so for your tangent you get: y=3x-2</p>
<p>for your normal, take opposite of the reciprocal of your original slope, so 3 turns to -1/3
y=(-1/3)x+b
using your point (1,1) again:
1=(-1/3)(1)+b, b=4/3
so y=(-1/3)x+(4/3)</p>
<p>Okay thank you. I am just cautious because one time I really thought I understood the math in pre-calc and all the problems were the same concept, so I wasted time doing them all wrong. Just didn't want that to happen again.</p>