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Fofino [41]
3 years ago
13

4. Find an equation of the line that is tangent to the graph of 4x + 2xy^2 +44 = y^3 at the point (-3,2).

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
6 0

Given:

The equation is:

4x+2xy^2+44=y^3

To find:

The equation of the tangent on the given equation at point (-3,2).

Solution:

We have,

4x+2xy^2+44=y^3

Differentiate with respect to x.

4(1)+2[x(2yy')+y^2(1)]+0=3y^2y'

4+4xyy'+2y^2=3y^2y'

4+2y^2=3y^2y'-4xyy'

4+2y^2=(3y^2-4xy)y'

Isolate y'.

\dfrac{4+2y^2}{3y^2-4xy}=y'

Now, we need to find the value of the derivative at point (-3,2).

y'_{(-3,2)}=\dfrac{4+2(2)^2}{3(2)^2-4(-3)(2)}

y'_{(-3,2)}=\dfrac{4+8}{12+24}

y'_{(-3,2)}=\dfrac{12}{36}

y'_{(-3,2)}=\dfrac{1}{3}

It means the slope of the tangent line is \dfrac{1}{3}.

The equation of tangent line that passes through the point (-3,2) with slope \dfrac{1}{3} is:

y-y_1=m(x-x_1)

y-2=\dfrac{1}{3}(x-(-3))

y-2=\dfrac{1}{3}x+1

y=\dfrac{1}{3}x+1+2

y=\dfrac{1}{3}x+3

Therefore, the equation of tangent line is y=\dfrac{1}{3}x+3.

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