Question

y=ax^3+bx^2+cx+5 touches the x-axis at the point (-2,0) then this cuts the y-axis at the point where its gradient is 3. then find the values of a,b and c.

Answer 1

⭕️Here the given equation is :-

y = ax³+ bx² + cx + 5

⭕️differenciate the equation.

y' = 3ax² + 2bx + c 

y'(0) = 3

0 + 0 + c = 3 

c = 3
 
⭕️ y(-2) = 0 so putting value:-

-8a + 4b -1 = 0➖➖➖➖➖(1) 

⭕️Also there's a slop at (2, 0)along x-axis:-

y'(-2) = 0 

12a - 4b + 3 = 0 ➖➖➖➖➖➖(2)

⭕️adding the equations  we get , 

 ⇒ 4a + 2 = 0 

 ⇒ a = -1/2

 ⇒4 + 4b -1 = 0 

 ⇒ 4b = -3 

 ⇒∴ b = -3/4 


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Answer 2

Answer :-

➡️Here the given equation is :-

y = ax³+ bx² + cx + 5

➡️differenciate the equation.

y' = 3ax² + 2bx + c 

y'(0) = 3

0 + 0 + c = 3 

c = 3

 

➡️ y(-2) = 0 so putting value:-

-8a + 4b -1 = 0➖➖➖➖➖(1) 

➡️Also there's a slop at (2, 0)along x-axis:-

y'(-2) = 0 

12a - 4b + 3 = 0 ➖➖➖➖➖➖(2)

➡️adding the equations  we get , 

 ⇒ 4a + 2 = 0 

 ⇒ a = -1/2

 ⇒4 + 4b -1 = 0 

 ⇒ 4b = -3 

 ⇒∴ b = -3/4