The question asks to find a cubic equation given point A (0/18) slope here is 0. Point B only given the x=20 and slope is -0.6.
Please help, been trying to solve it for so long now. :(
1.) y=ax^3+bx^2+cx+d
2.) y'=3ax^2+2bx+c
-0,6=1200a+40b+1c
18=0a+0b+0c+1d
On
You have too little information, but the general method is:
Write the general equation for a cubic, differentiate it, and plug in the information you have in those two expressions. That will give you a set of equations to solve.
But the information you mention will only give three equations with four unknonwns.
And you haven't used that the slope at $0$ is $0$.
What you wrote is not very clear. But, if I understand it correctly, you are only given three pieces of information -- one function value and two slopes. From this, you won't be able to calculate unique values for the four unknowns $a,b,c,d$.
But, as the comment from @Justin points out, you could find some values of $a,b,c,d$ that will work: \begin{align} y(0) &= 18 \; \Rightarrow \; d = 18 \\ y'(0) &= 0 \; \Rightarrow \; c = 0 \\ y'(20) &= -0.6 \; \Rightarrow \; 1200a+40b+c = -0.6 \end{align} From the third equation, we get $1200a+40b = -0.6$. We can just set $a=1$, arbitrarily, and then we get $b = -1200.6/40 = -30.015$.
So, a suitable cubic (but not the only one) is: $$ y = x^3 -30.015x^2 + 18 $$