Question

Given \[y^{2} = x^{2}(x+7)\]
Find an equation for the tangent line at ( -3, 6).
Find an equation for the normal line at ( -3, 6).
Whats the difference? How do I find each? Confused. Help?

Answer 1

tangent line is line that touches the point on the curve
normal line is line perpendicular to the tangent line
to find tangent line differentiate the function to find slope
slope of normal is the opposite reciprocal of that of the tangent

Answer 2

Can you help me solve the problem? Itried to work it out, but it looked complicated. :/

Answer 3

equation for a line tangent is mTan(x-x1)=y-y1
equation for a normal line is -(1/mTan)(x-x1)=y-y1
mtan=derivative since it touches a point once

Answer 4

\[y=\sqrt{x ^{2}(x+7)}\]
differentiate using chain rule and product rule
u = x^2(x+7)
du = 2x(x+7) + x^2
derivitive of sqrt(u) = 1/2sqrt(u)
\[y' = [x ^{2}+2x(x+7)]/2\sqrt{x ^{2}(x+7)}\]

Answer 5

slope of normal is opposite reciprocal of y'
\[slopeNormal =-2\sqrt{x ^{2}(x+7)}/[x ^{2}+2x(x+7)]\]

Answer 6

plug in -3 for x
y' = -5/4
slopeN = 4/5

Tangent:
6 = -5/4(-3) +b
b = 9/4
y = -5/4 x + 9/4

Normal:
6=4/5(-3) +b
b = 42/5
y = 4/5 x + 42/5