1) Find the points on the graph of y^2 = x^3 - 3x + 1 a) First show that 2yy\' =

Question

1) Find the points on the graph of y^2 = x^3 - 3x + 1

a) First show that 2yy' = 3x^2 -3 where y' = dy/dx

b) Do not solve for y'. Rather, set y' = 0 and solve for x. This yields two values for x where the slope may be zero.

c) Show that positive value of x does not correspond to a point on the graph.

d) The negative value corresponds to the two points on the graph where the tangent line is horizontal. Find their coordinates

2) Find all points on the graph of 3x^2 + 4y^2 + 3xy = 24 where the tangent line is horizontal.

3) (x+2)^2 - 6(2y +3)^2 = 3, (1, -1) Find the dy/dx at the given point.

Explanation / Answer

2)

3x2 + 4y2 + 3xy =24

derivative both sides:

6x + 8y y' + 3y + 3x y' = 0

y' = (-6x-3y) / (8y+3x)

let y' = 0

= (-6x-3y) / (8y+3x) = 0

= -6x - 3y = 0

y = 2x

when y = 2x, the function have a horizontal tangent line

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