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Pondera
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Homework Statement
Find an equation of the tangent line to this curve at the point (1, -2).
Homework Equations
The Attempt at a Solution
2y' = 3x^2+6x
y' = 3x^2+6x
y'=3/2x^2+3x
y+2=3(x-1)
y+2=3x-3
y=3x-5
Last edited:
where did the 2 go?Pondera said:Homework Statement
Homework Equations
The Attempt at a Solution
2y' = 3x^2+6x
y' = 3x^2+6x
where did the power of x go?Pondera said:y'=3/2x+3x
Pondera said:y+2=3(x-1)
y+2=3x-3
y=3x-5
Implicit differentiation is a method used in calculus to find the derivative of an equation that is not written in the form y = f(x). It allows us to find the slope of a tangent line at a given point on a curve.
To find the derivative using implicit differentiation, we treat the dependent variable (usually y) as a function of the independent variable (usually x) and differentiate both sides of the equation with respect to x. This allows us to isolate the derivative and solve for it.
The equation for a tangent line at a point (x0, y0) on a curve is y - y0 = m(x - x0), where m is the slope of the tangent line. Using implicit differentiation, we can find the slope at a given point and then substitute it into this equation to find the equation of the tangent line.
A horizontal tangent is a line that is parallel to the x-axis and has a slope of 0. This means that the derivative of the function at that point is equal to 0. In other words, the function has no change (or rate of change) at that point.
To find horizontal tangents using implicit differentiation, we set the derivative equal to 0 and solve for the x-values that make the derivative 0. These x-values correspond to the points where the function has a horizontal tangent.