find the equation of the curve for which y'''=24x-6 if the curve passes through (-1,8) and is tangent to the line y=4x at (1,4).

help please thanks :-)

y"' = 24x-6

y" = 12x^2-6x+2a
y' = 4x^3-3x^2+2ax+b
y = x^4-x^3+ax^2+bx+c

Now what else do we know?
y(-1) = 8
y(1) = 4
y'(1) = 4
So,
1+1+a-b+c = 8
1-1+a+b+c = 4
4-3+2a+b = 4
Solve for a,b,c and you have

y = x^4-x^3+2x^2-x+3

I'll let you verify that it meets the conditions.

To find the equation of the curve, we need to integrate the given expression for y''' with respect to x three times. This will allow us to obtain an expression for y in terms of x.

Let's start by integrating y''' = 24x - 6 with respect to x.

∫(y''') dx = ∫(24x - 6) dx

Integrating, we get:

y'' = 12x^2 - 6x + C1

Now, we integrate y'' with respect to x:

∫(y'') dx = ∫(12x^2 - 6x + C1) dx

Integrating, we get:

y' = 4x^3 - 3x^2 + C1x + C2

Finally, we integrate y' with respect to x:

∫(y') dx = ∫(4x^3 - 3x^2 + C1x + C2) dx

Integrating, we get:

y = x^4 - x^3 + (C1/2)x^2 + C2x + C3

Now that we have the general equation for the curve, we need to use the given conditions to determine the specific values of C1, C2, and C3.

We are given that the curve passes through the point (-1, 8). Substituting these values into the equation, we get:

8 = (-1)^4 - (-1)^3 + (C1/2)(-1)^2 + C2(-1) + C3
8 = 1 + (-1) + (C1/2) + (-C2) + C3
8 = 2 + (C1/2) - C2 + C3

Simplifying, we have:

(C1/2) - C2 + C3 = 6 (Equation 1)

We are also given that the curve is tangent to the line y = 4x at the point (1, 4). Since the curve is tangent to the line at that point, it means that the slopes of both the line and the curve are equal at that point.

The slope of the line y = 4x is 4. Substituting these values into the derivative of the equation of the curve, we get:

y' = 4(1)^3 - 3(1)^2 + C1(1) + C2
4 = 4 - 3 + C1 + C2
1 = C1 + C2 (Equation 2)

Now, we have two equations (Equation 1 and Equation 2) with two unknowns (C1 and C2). We can solve these equations simultaneously to find the values of C1 and C2.

Subtracting Equation 2 from Equation 1, we get:

(C1/2) - C2 + C3 - (C1 + C2) = 6 - 1
(C1/2) - C2 + C3 - C1 - C2 = 5
-C1/2 - 2C2 + C3 = 5 (Equation 3)

From Equation 2, we have:

C1 + C2 = 1

Multiplying this equation by 2, we get:

2C1 + 2C2 = 2 (Equation 4)

Adding Equation 4 to Equation 3, we eliminate C1 and C2:

2C1 + 2C2 - C1/2 - 2C2 + C3 = 2 + 5
(4C1 - C1)/2 + (4C2 - 2C2) + C3 = 7
3C1/2 + 2C2 + C3 = 7
3C1 + 4C2 + 2C3 = 14 (Equation 5)

Now, we can solve Equation 5 simultaneously with Equation 1 to find the values of C1, C2, and C3.

Solving Equations 1 and 5 simultaneously can be done through various methods, such as substitution or elimination. Once you obtain the values of C1, C2, and C3, you can substitute them back into the general equation y = x^4 - x^3 + (C1/2)x^2 + C2x + C3 to obtain the specific equation for the curve.