First Integral

The first integration gives us the second derivative:

y’’ x^4 / 2 C1

Second Integral

Next, we integrate again to find the first derivative:

y’ x^5 / 10 C1x C2

Third Integral

y x^6 / 60 C1x^2 / 10 C2x C3

Here, C1, C2, and C3 are the constants of integration.

Applying the Given Conditions

Now, we need to apply the given conditions to find the specific values of C1, C2, and C3.

Tangent Slope Condition

We are given that the slope of the inflectional tangent at x -1 is 2/3 of its ordinate. The slope is given by y’, and the ordinate is given by y. At x -1, we have:

y(-1) (-1)^6 / 60 C1(-1)^2 / 10 C2(-1) C3

y’(-1) (-1)^5 / 10 C1(-1) C2

Given that y’(-1) 2 and y(-1) 2/3 * y’(-1) 4/3, we can set up the following equations:

-1/10 C1 - C2 2

1/60 - C1 - C2 - C3 4/3

Inflection Point Condition

The inflection point is where the second derivative is zero, i.e., y’’(-1) 0. From the second integration, we have:

y’’ x^3 / 2 C1

Setting x -1 and y’’(-1) 0, we get:

-1 / 2 C1 0 rarr; C1 1 / 2

Substituting C1 Back

Substituting C1 1 / 2 into the previous equation, we get:

-1 / 10 - 1 / 4 - C2 2

Clean up the equation:

C2 -1/10 - 1/4 - 2 -9/10

Final Constants

Now, substituting C1 1 / 2 and C2 -9 / 10 into the third equation:

-1/60 1/20 9/10 C3 4/3

C3 4/3 - 1/60 1/20 - 9/10 4/3 - 1/60 3/60 - 54/60 80/60 - 58/60 22/60 11/30

Final Equation

Substituting all these constants, we get the final equation of the curve:

y x^6 / 60 - x^5 / 48 9x^4 / 10 11x / 30

Thus, the equation of the curve is y x^6 / 60 - x^5 / 48 9x^4 / 10 11x / 30.

Conclusion

By solving the given conditions step-by-step, we have derived the equation of the curve. This process involves understanding the third derivative, integrating, and applying specific slope and inflection point conditions. The final equation is a complex polynomial that accurately meets the given requirements.