The curve y=ax³+bx²+c passes through the points P(-1,0) and Q(0,5)?

May 2021 2 95 Report

The curve y=ax³+bx²+c passes through the points P(-1,0) and Q(0,5)?

and the tangent to the curve at the point R(1,4) is parallel to the x-axis.

(a) Find the values of a,b and c. (4 marks)

(b) Hence, show that the point S(0,5) is the maximum point of the curve. (3marks)

help me to solve this additional mathematics question . thank you . God bless you !

First U will SubStitute in 2 points P and Q to get 2 equations and solve them simultaneously

on Substituting with P x = -1 and y = 0 Function will be -a + b + c = 0 => Equ(1)

the second is Q x = 0 y = 5 on Sub. u get 5 = 0 + 0 + c So c = b => Equ(2)

Substitute in one we get that b - a = -5

Then we get the equation of tangent of this curve and it is equal Derivative of it

dy/dx = 3ax^2 + 2bx

Since it pass thorough R Then Substitute in dy/dx to get another equation

R x = 1

So dy/dx = 3a + 2b

Since tangent is parallel to x-axis then its slope = zero

and Slope of tangent = the Derivative of the function

Then 3a + 2b = 0 => Equ(3)

Solve it with Equation (1)

b - a = -5 (Times2)

2b - 2a = -10

3a + 2b = 0

Subtract both from each other

We get -5a = -10 then a = 2 by substituting u get b = -3 So a = 2 , b= -3 c = 5

(b)Didnot Learn It Till Nw Sry i cant help for number b :S