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anonymous

  • 5 years ago

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?

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  1. dumbcow
    • 5 years ago
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    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

  2. anonymous
    • 5 years ago
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    Can you help me solve the problem? Itried to work it out, but it looked complicated. :/

  3. anonymous
    • 5 years ago
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    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

  4. dumbcow
    • 5 years ago
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    \[y=\sqrt{x ^{2}(x+7)}\] 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) \[y' = [x ^{2}+2x(x+7)]/2\sqrt{x ^{2}(x+7)}\]

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

  6. dumbcow
    • 5 years ago
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    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

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