The curve below is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the point (-2, 1/5). y = How do I do this? Please help me out It looks so weird
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Originally Posted by miss.strw The curve below is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the point (-2, 1/5). slope, $\displaystyle m = y'(-2)$ now use the point-slope form of a linear equation ... $\displaystyle y - y_1 = m(x - x_1)$ where $\displaystyle (x_1,y_1)$ is the given point.
Originally Posted by skeeter slope, $\displaystyle m = y'(-2)$ now use the point-slope form of a linear equation ... $\displaystyle y - y_1 = m(x - x_1)$ where $\displaystyle (x_1,y_1)$ is the given point. Do i need to plug in -2 into the equation to get m? or u're telling me that m = -2? thanks for helpin! please reply soon! i do not know how to get the y` like i don't know how to do tat.
Originally Posted by miss.strw Do i need to plug in -2 into the equation to get m? or u're telling me that m = -2? thanks for helpin! please reply soon! i do not know how to get the y` like i don't know how to do tat. the slope is the derivative of the function evaluated at x = -2 what does this mean? ... like i don't know how to do tat. you don't know how to find a derivative?
Originally Posted by skeeter the slope is the derivative of the function evaluated at x = -2 what does this mean? ... you don't know how to find a derivative? Not for those one. i don't know is it like 1/1+x^2 = 2x? so it's like y-(1/5)=(2x)(x-(-2)) ? thank you so much for helping!
Originally Posted by miss.strw Not for those one. i don't know is it like 1/1+x^2 = 2x? no so it's like y-(1/5)=(2x)(x-(-2)) ? no $\displaystyle y = \frac{1}{1+x^2}$ use the quotient rule to find y' , or ... $\displaystyle y = (1+x^2)^{-1}$ ... take the derivative using the chain rule.
Originally Posted by skeeter $\displaystyle y = \frac{1}{1+x^2}$ use the quotient rule to find y' , or ... $\displaystyle y = (1+x^2)^{-1}$ ... take the derivative using the chain rule. g`(-2)h(-2)-g(-2)h`(-2) --------------------------- [h(-2)]^2 this doesn't make sense to me if i do the derivative 1 ------- (1+x^2) is derivative is x^2 = 2x
you need to find the derivative first, then evaluate it at x = -2
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