Finding the Normal Line for a Curve at a Given Point

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To find the normal line for the curve y = -3x^2 at the point (-1, -3), first calculate the derivative to determine the slope of the tangent line, which is found to be 6x + 3. At x = -1, the slope of the tangent line is 3. The slope of the normal line, being perpendicular to the tangent, is the negative reciprocal, resulting in a slope of -1/3. Using the point-slope form of a line with the point (-1, -3), the equation of the normal line is derived as y = -1/3x - 8/3. The final equation for the normal line is y = -1/6x - 19/6.
fatima_a
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find the equation of the normal line to the curve y = -3x^2 at the point (-1,-3).


you have to use derivatives.


i am not asking for the equation of the tangent line i know that is 6x + 3. please show all the steps.

the answer is y = -1/6x - 19/6
 
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Do you know the relationship between the gradient of two perpendicular lines?
 
Another term for gradient in this context is "slope." You can use calculus to find the slope of the tangent line. What is the slope of the line perpendicular to the tangent line? That will be the slope of the line you want. You know a point on this line, so you will have all the information you need to get the equation of this second line.
 
thanks i got it
 

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