Tangents... 唔識......

Question 1.
y =ax^2+bx+c
1 = a(1)^2 +b(1) + c
1 = a +b + c ------------- (1)

3 = a(3)^2 +b(3) + c
3 = 9a +3b + c ---------- (2)

dy/dx = 2ax + b
7 = 2a(3) + b
7 = 6a + b
b = 7 – 6a -------------- (3)

Substitute (3) into (1) and (2)
1 = a + (7 – 6a) + c
-6 = -5a + c -------------- (4)

3 = 9a + 3(7 – 6a) + c
3 = 9a +21 – 18a + c
-18 = -9a + c ------------- (5)
Equation (4) – equation (5)
12 = 4a
a = 12/4 = 3
substitute a = 3 into equation (4)
-6 = -5(3) + c
c = 15 – 6 = 9
From equation (1)
a + b + c = 1
4 + b + 9 = 1
b = -12

a = 3, b = -12, c + 9


Question 2:
y =ax^2+bx+c
3 =a(1)^2+b(1)+c
a + b + c = 3 -------------------- (1)

1 =a(2)^2+b(2)+c
1 = 4a + 2b + c ------------------- (2)

dy/dx = 2ax + b
-4 = 2a(2) + b
-4 = 4a + b
b = -4a – 4 -------------------- (3)
Substitute equation (3) into equatios (1) and (2)
a + (-4a -4) + c = 3
-3a + c = 7 --------------------------(4)

1 = 4a + 2(-4a – 4) + c
1 = 4a -8a – 8 + c
-4a + c = 9 --------------------- (5)
(4) – (5)
a = - 2
b = -4a – 4 -------------------- (3)
b = -4(-2) – 4
b = 8 – 4 = 4
From Equation (5)
c = 4a + 9
c = 4(-2) + 9 = 1

a = -2, b = 4, c = 1


Question 3 is very similar to question 1 and question 4.
 Run out of space soon. Better do question (4)


Question 4.

y = ax^3+bx^2+cx +d
1 = a(0)^3+b(0)^2+c(0) +d
d = 1

4 = a(3)^3+b(3)^2+c(3) +1
27a + 9b +3c + 1 = 4
27a + 9b +3c = 3 ---------------- (1)

dy/dx = 3ax^2 + 2bx + c
1 = 3a(0)^2 + 2b(0) + c
c = 1

27a + 9b +3c = 3
27a + 9b +3(1) = 3
27a + 9b = 0 ------------------ (2)

dy/dx = 3ax^2 + 2bx + 1
-2 = 3a(3)^2 + 2b(3) + 1
-2 = 27a + 6b + 1
27a + 6b = -3 ------------------ (3)
(2) – (3)
3b = 3
b = 1
27a + 9(1) = 0
27a = -9
a = -1/3

a = -1/3, b =1, c =1, d = 1

Comment:
You are the one who asked about the limit and derivative recently if I remember right. Please try do question (3). Good luck!