The tangent line to y=f(x)y=f(x) at (−3,−2)(−3,−2) passes through the point (−4,−3)(−4,−3). Compute the following. f(-3)= f'(-3)=

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A: The given function is: f(x,y)=y2x3-y Now, the function is defined for all y but if  x3-y≥0⇒x3≥y⇒x≥y3…

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A:

Q: Find an equation for the line tangent to the graph of 7x f(x) = x + 7 at the point (3, 2.1). y =

A: Given curve is Differentiating with respect to x, we get

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A:

Q: Find an equation for the line that is tangent to the curve y=x3 −9x at the point (3,0).

A: Given: The given curve is y=x3-9x and the point 3,0. To find: The main aim is to find the equation…

Q: Find y as a function of a if y(0) = -2, (0) = 9, '(0) = -62. y(x) = y" - 4y" -y + 4y = 0,

A: Given that y''' -4y'' - y' + 4y = 0 ,   y(0) = - 2 y(0) = 9 '(0) = -62.

Q: Find y as a function of a if a?y" + 13ry' + 36y = a, %3D y(1) = -2, y'(1) = -1. y =

A: