**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 P_{0} = (0,0) and P_{3} = (-4,2)

Whereas the curve is controlled by P_{1} (14,10) and P_{2} (4,0)

Number of control points,

P_{0}=(0,0) , P_{1}=(14,10) , P_{2}=(4,0) , P_{3}=(-4,2)

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

**B(t) = P _{0}B_{0,3}(t)+ P_{1}B_{1,3}(t) + P_{2}B_{2,3}(t)+ P_{3}B_{3,3}(t)**

Where, **B _{k,n}(t)= ^{n}C_{k }u^{k} (1-t)^{n-k}**

**B _{0,3}(t)**=

^{3}C

_{0 }t

^{0}(1-t)

^{3-0}

= 1(1-t)^{ 3}

^{ }= 1(1-t)^{ 3}

^{ }= **(1-t) ^{3}**

**B _{1,3}(t)**=

^{3}C

_{1 }u

^{1}(1-t)

^{3-1}

= t(1-t)^{ 2}

^{ }= t(1-t)^{ 2}

^{ }=** 3t (1-t) ^{2}**

**B _{2,3}(t)**=

^{3}C

_{2 }t

^{2}(1-t)

^{3-2}

= t^{2}(1-t)^{ 1}

^{ }= t^{2}(1-t)^{ 1}

^{ }= **3t ^{2} (1-t)^{1}**

^{ }

B_{3,3}(t)= ^{3}C_{3 }u^{3} (1-t)^{3-3}

= t^{3}(1-t)^{ 0}

^{ }= t^{3}(1-t)^{ 0}

^{ }= ** t ^{3}**

B(t)= P_{0 }(1-t)^{3} + P_{1 }3t (1-t)^{2}+ P_{2 }3t^{2} (1-t)^{1}+ P_{3} t^{3}

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

P_{0}=(0,0) , P_{1}=(14,10) , P_{2}=(4,0) , P_{3}=(-4,2)

= 0(1-t)^{3} + (14) _{ }3t (1-t)^{2}+ (4) 3t^{2} (1-t)+ (-4) t^{3}

^{= }42t(t^{2}+1-2t)+ 12t^{2}(1-t)-4t^{3}

= 42t^{3}+42t-84t^{2}+12t^{2}-12t^{3}-4t^{3}

**=26t ^{3}-72t^{2}+42t**

Using this equation lets calculate equation for y control points

= 0(1-t)^{3} + (10) _{ }3t (1-t)^{2}+ (0) 3t^{2} (1-t)+ (2) t^{3}

=30t (1+t^{2}-2t)+2t^{3}

=30t+30t^{3}-60t^{2}+2t^{3}

= **32t ^{3} -60t^{2}+30t**

**So, The answer is P(t) =[ x(t) y(t)] =[ (26t ^{3}-72t^{2}+42t) (32t^{3} -60t^{2}+30t)]**