Solution:

B(t) = ∑^k_i=0   P_k B_k,n(t)         where, 0<=t<=1

It is given that curve passes (0,0) and (-4,2), means starting point of the curve is P₀ = (0,0) and P₃ = (-4,2)

Whereas the curve is controlled by P₁ (14,10) and P₂ (4,0)

Number of control points,

P₀=(0,0) ,  P₁=(14,10) , P₂=(4,0)  , P₃=(-4,2)

Degree of equation n =Control_point -1 = 4 -1 = 3

B(t) = P₀B_0,3(t)+ P₁B_1,3(t) + P₂B_2,3(t)+ P₃B_3,3(t)

Where,  B_k,n(t)=  ⁿC_k    u^k (1-t)^n-k

B_0,3(t)=  ³C₀    t⁰ (1-t)^3-0

=   1(1-t) ³

               =   1(1-t) ³

             =  (1-t)³

B_1,3(t)=  ³C₁    u¹ (1-t)^3-1

=   t(1-t) ²

               =   t(1-t) ²

             =  3t (1-t)²

B_2,3(t)=  ³C₂    t² (1-t)^3-2

=   t²(1-t) ¹

               =   t²(1-t) ¹

             =  3t² (1-t)¹

B_3,3(t)=  ³C₃   u³ (1-t)^3-3

=   t³(1-t) ⁰

               =   t³(1-t) ⁰

             =  t³

B(t)=      P₀ (1-t)³ + P₁ 3t (1-t)²+ P₂ 3t² (1-t)¹+ P₃ t³

Now, Using this equation lets calculate equation for x control point

P₀=(0,0) ,  P₁=(14,10) , P₂=(4,0)  , P₃=(-4,2)

= 0(1-t)³ + (14)  3t (1-t)²+ (4) 3t² (1-t)+ (-4) t³

⁼ 42t(t²+1-2t)+ 12t²(1-t)-4t³

= 42t³+42t-84t²+12t²-12t³-4t³

=26t³-72t²+42t

Using this equation lets calculate equation for y control points

= 0(1-t)³ + (10)  3t (1-t)²+ (0) 3t² (1-t)+ (2) t³

=30t (1+t²-2t)+2t³

=30t+30t³-60t²+2t³

= 32t³ -60t²+30t

So, The answer is P(t) =[ x(t) y(t)] =[ (26t³-72t²+42t) (32t³ -60t²+30t)]