91.7k views
2 votes
X³+ax²-4x+b has a stationary point at P(2,3) find a and b

1 Answer

6 votes

Final answer:

The values of a and b are -2 and 11, respectively, which can be determined by setting up equations using the fact that P(2,3) is a stationary point on the curve represented by the equation x³+ax²-4x+b.

Step-by-step explanation:

To find the values of a and b in the cubic equation x³+ax²-4x+b given the stationary point P(2,3), we need to use two main concepts: the coordinates of P tell us the value of the function and its derivative at x=2.

First, since P(2,3) is a point on the curve, when x=2 the value of the function is 3. Thus, plugging these into the equation we get:

2³ + 2²a - 4(2) + b = 3

This simplifies to:

8 + 4a - 8 + b = 3

4a + b = 3

Next, since P(2,3) is a stationary point, the derivative of the function at x=2 is 0. The derivative of the function is:

3x² + 2ax - 4

Setting x=2 and equating to zero gives us:

3(2²) + 2a(2) - 4 = 0

12 + 4a - 4 = 0

4a = -8

a = -2

Having found the value of a, we can substitute it back into the equation 4a + b = 3:

4(-2) + b = 3

-8 + b = 3

b = 11

Therefore, the values are a = -2 and b = 11.

User Squiggle
by
7.6k points