The curve y=2x^3+ax^2+bx-30 has a stationary point when x=3. The curve passes through the point (4,2). (A) Find the value of a and the value of b.

#secondderivative #stationarypoints

A stationary point at x = 3 means the derivative dy/dx = 0 at that point. Differentiating, we have

dy/dx = 6x ² + 2ax + b

and so when x = 3,

0 = 54 + 6a + b

or

6a + b = -54 … … … [eq1]

The curve passes through the point (4, 2), which is to say y = 2 when x = 4. So we also have

2 = 128 + 16a + 4b - 30

or

16a + 4b = -96

4a + b = -24 … … … [eq2]

Eliminate b by subtracting [eq2] from [eq1] and solve for a, then for b :

(6a + b) - (4a + b) = -54 - (-24)

2a = -30

a = -15   ===>   b = 96

on a coordinate grid, point a is at (−4.0, −2.4) and point b is at (4.0, −2.4). point b is a reflection of point a across the –axis. input either a lowercase x or i think is the right awnser

step-by-step explanation:

, and

for

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

note the following:

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

, and for