solve cubic curve with 4 unkown values which have gradients -7 and 0 at (0,3) and (1,0) respectively.

Question

^{y=ax3 +bx2 +cx+d}

the curve passes through points (0,3) and (1,0) with gradients -7 and 0 respectively. find the values of a,b,c and d

Answer 1

Interesting problem. Since (0,3) lies on the graph it is a solution so a(0)^3 + b(0)^2 +c(0) +d =3 so d=3

The gradient or slope is the same thing as the derivative of the equation . The derivative is 3ax^2 +2bx +c

so at the point (0,3) the derivative is -7 so 3a(0)^2 +2b(0) +c = -6 so c=-7

At the point (1.0) the derivative is 0 so 3a(1)^3 +2b(1) + c = 0 and we know c=-7 so 3a + 2b =7 and

0 = a(1)^3 + b(1)^2 +c(1) +d we know c and d so

0= a+b -7 +3 this gives a+b =4 Now we have 2 equations with 2 unknowns which allows us to solve for a and b

1) a+b =4

2) 3a + 2b= 7

multiply equation 1 by 2 gives us 2a +2b =8

3a + 2b =7

Subtract equation 1 from equation 2 gives us a = -1 so b=5

So our equation is y = -x^3 +5x^2 -7x +3

Plot this in your graphing calculator and you will see it satisfies the parameters given