# At every point of the graph of the function y=f(x) we have y''=30x and at the point (0,3), the tangent to the graph of y = f(x) is parallel to the line 9x-4y+4 = 0. Find f(x).

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The second derivative of the function y = f(x) is `y'' = 30*x`

The derivative of a function f(x) at a point x = a is equal to the slope of the tangent to the graph of y = f(x) at x = a. Here, the tangent is parallel to the line 9x - 4y + 4 = 0

9x - 4y + 4 = 0

=> 4y = 9x + 4

=> y = (9/4)x + 1

The slope of the tangent is (9/4).

At x = 0, f'(x) = 9/4

`f'(x) = int f''(x) dx`

=> `int 30x dx`

=> 15x^2 + C``

`f'(0) = 9/4`

=> `C = 9/4`

`f'(x) = 15x^2 + 9/4`

`f(x) = int f'(x) dx`

=> `int 15x^2 + 9/4 dx`

=> `15x^3/3 + (9/4)*x + C`

As the point (0, 3) lies on the graph of f(x)

3 = 0 + 0 + C

**This gives **`f(x) = 5x^3 + (9x)/4 + 3`

Hi,

The f(x) will be after intregation: 5x^3+9x/4+3