Given the curve y= x^3 + 2x^2 + tx + 3

First we will determine the tangent line.

Since the tangent line is horizontal, then the slope is 0.

Now we need to find the point of tangent.

We will determine the first derivative.

==> y' = 3x^2 + 4x + t

==> 3x^2 + 4x + t = 0

Since the curve has only one tangent line, then the quadratic equation has one real solution. Therefore, the discriminant is 0.

==> b^2 - 4ac = 0

==> a= 3 b= 4 c = t

==> 16 - 4*3*t = 0

==> 16 - 12t = 0

==> 12t = 16

==> t= 16/12 = 4/3

==> **t= 4/3.**

==> Now we will substitute into y'.

==> y' = 3x^2 + 4x + 4/3 = 0

==> x1= -4 + sqrt(b62-4ac)/ 2*3 = -4/6 = -2/3

**==> x = -2/3**

**Therefore t = 4/3 and x = -2/3**

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