Question #4828

find two straight lines tangent to y=x^3 and they pass through point (2,8)

Expert's answer

The equation of a tangent line at point a is

f(x) = y + y'(x-a) = a^3 + 3a^2(x-a).

Substituting point (2,8) we obtain:

8 = a^3 + 3a^2(2-a).

Modifying this equation,

-2a^3 + 6a^2 = 8;

a^3 - 3a^2 + 4 = 0.

We need to factorize left-handed side of the equation. It can be easily seen that a=-1 is a solution of this equation. Dividing by a+1 we obtain

(a+1)(a^2-4a+4) = 0,

or

(a+1)(a-2)^2 = 0.

So, there are two solutions: a=-1 and a=2.

At last, we have

f(x) = (-1)^3 + 3(-1)^2(x-(-1)) = -1+3(x+1) = 3x+2 - first tangent line, and

f(x) = 2^3 + 3*2^2*(x-2) = 8 + 12(x-2) = 12x-16 - second tangent line.

f(x) = y + y'(x-a) = a^3 + 3a^2(x-a).

Substituting point (2,8) we obtain:

8 = a^3 + 3a^2(2-a).

Modifying this equation,

-2a^3 + 6a^2 = 8;

a^3 - 3a^2 + 4 = 0.

We need to factorize left-handed side of the equation. It can be easily seen that a=-1 is a solution of this equation. Dividing by a+1 we obtain

(a+1)(a^2-4a+4) = 0,

or

(a+1)(a-2)^2 = 0.

So, there are two solutions: a=-1 and a=2.

At last, we have

f(x) = (-1)^3 + 3(-1)^2(x-(-1)) = -1+3(x+1) = 3x+2 - first tangent line, and

f(x) = 2^3 + 3*2^2*(x-2) = 8 + 12(x-2) = 12x-16 - second tangent line.

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