# What is the curve which has a slope of 6x^2 at (x, y) and if the curve passes through (4,8)?

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We know that the slope of a curve y = f(x) at the point (x, y) is given as dy/dx.

Now, we are given that dy/dx = 6x^2. We will solve this equation by integrating both sides with respect to x:

dy/dx= 6x^2

=> y = Int [6x^2 dx]

= 2x^3 +C

Now to identify the particular solution whose graph passes through the point (4, 8), we find the value of C for which y= 8 when x = 4

y = 2x^3 +C

=> 8 = 2*4^3 +C

=> C = -128 + 8

=> C= -120

**Therefore the curve we are looking for is y = 2x^3 – 120.**

Given the function of the slope is f(x) = 6x^2

Then, we know that the curve that has a slope of 6x^2 is integral f(x).

==> Let the curve be F(x) such that:

F(x) = intg f(x) dx

==> F(x) = intg (6x^2 ) dx

= 6x^3/ 3 + C

==> F(x) = 2x^3 + C

But the curve passes through the point ( 4, 8).

Then the point (4,8) should verify the equation of the curve.

==> F(4) = 8

==> 2(4^3) + C = 8

==> 128 + C = 8

==> B = -120.

==>** F(x) = 2x^3 - 120**

the slope of a curve is given by f'(x).

Therfore f'(x) = 6x^2.

Therefore the curve f(x) = Int f'(x)dx .

f(x) = Int 6x^2 dx = (1/2+1)6x^(2+1) +C = (6/3) x^3 +C.

f(x) = 2x^3+C.

Since f(x) passes therough (4,8) ,

f(4) = 8 , or 2*4^3+C = 8. So 128+C = 8, Or C = 8-128 = -120.

Therefore we rewite with C = -120:

f(x) = 2x^3-128.

Therefore the required curve is f(x ) = 2x^3 -120.