# The curve for which dy/dx = 3x^2+ax+b, where a and b are constants, has stationary points at (1,0) , (-3,32). Find the values of a and b

## To find the values of a and b, we can use the information provided about the stationary points.

First, let's find the derivative of 3x^2 + ax + b with respect to x:

dy/dx = 6x + a

Now, using the fact that the curve has a stationary point at (1,0), we can substitute x = 1 into the derivative and set it equal to 0:

6(1) + a = 0

6 + a = 0

a = -6

To find the value of b, we can substitute the coordinates of the other stationary point (-3, 32) into the original equation:

0 = 3(-3)^2 + (-6)(-3) + b

0 = 3(9) + 18 + b

0 = 27 + 18 + b

0 = 45 + b

b = -45

Therefore, the values of a and b are a = -6 and b = -45.

## To find the values of a and b, we need to use the given information about the stationary points of the curve.

Since the stationary points are at (1,0) and (-3,32), we know that the slope of the curve at these points is zero. In other words, dy/dx equals zero at these points.

Using the given equation, we can set up two equations by plugging in the x-coordinates of the stationary points and setting the equation equal to zero:

For the point (1,0):

0 = 3(1)^2 + a(1) + b

0 = 3 + a + b ----(1)

For the point (-3,32):

0 = 3(-3)^2 + a(-3) + b

0 = 27 - 3a + b ----(2)

Now we have a system of two equations with two unknowns (a and b). We can solve this system of equations to find the values of a and b.

To do this, we can subtract equation (2) from equation (1) to eliminate b:

0 - (27 - 3a + b) = 3 + a + b - (27 - 3a + b)

0 - 27 + 3a - b = 3 + a + b - 27 + 3a - b

-27 + 3a - b = 3 + a + 3a

Combining like terms:

4a - 2b - 27 = 3 + 4a

Now, we can simplify further:

4a - 2b - 27 = 3 + 4a

By subtracting 4a from both sides and adding 2b to both sides:

-2b - 27 = 3

By adding 27 to both sides:

-2b = 30

Finally, by dividing both sides by -2, we find:

b = -15

Now that we have the value of b, we can substitute it back into equation (1) to find the value of a:

0 = 3 + a + b

0 = 3 + a + (-15)

0 = 3 + a - 15

0 = -12 + a

By adding 12 to both sides:

12 = a

Therefore, we have found that a = 12 and b = -15.

## at each starionary point, dy/dx = 0, so when dy/dx = 3x^2 + ax + b

at (1,0) , 0 = 0 + 0 + b

b = 0

at (-3,32), 0 = 27 - 3a + 0

a = 9

a = 9, b = 0

check:

dy/dx = 3x^2 + 9x

at stationary points, 3x^2 + 9x = 0

3x(x+3) = 0

such points have x = 0 or x = -3, as shown by the given points.