Develop Cubic equation with a few points

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To develop the cubic equation, the function must have a y-intercept at 1, a local minimum at (3,1), and pass through the point (2,5). The equation can be structured as f(x)=a*x^3+b*x^2+c*x+1, with the known value of d=1. To find the coefficients a, b, and c, two equations can be derived from the points (3,1) and (2,5), but a third equation is needed, which relates to the local minimum condition. The local minimum indicates that the derivative at x=3 must equal zero, providing the necessary third equation. Understanding this relationship is crucial, especially for those who have not yet studied calculus.
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Homework Statement



Write the equation of the cubic function with a y-intercept at 1, a local minimum of (3,1) and through point (2,5)


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The Attempt at a Solution


I know that d= +1 and that's about it. I am really stuck and needs to know this for my summative on Tuesday. If anyone has an idea what to do, pleasseee help.
 
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Ok, yes d=1. That's a start. So f(x)=a*x^3+b*x^2+c*x+1. You need to find a,b and c. You know f(3)=1 and f(2)=5, right? That's 2 equations in the 3 unknowns a, b and c. You need one more. What does having a local min at x=3 tell you?
 
a local min means that the slope is zero...does that mean that y=0?
I still don't understand what to do...do I use elimation with the 3 equations?
 
LSCupcake said:
a local min means that the slope is zero...does that mean that y=0?
I still don't understand what to do...do I use elimation with the 3 equations?

Yes, you are going to use elimination with the 3 equations once you get them. A local min means slope is zero, alright, but that doesn't mean y=0, it means f'(3)=0. What's f'(x)?
 
I can't use f'(x) because we learn that in calculas, that is a derivative right? And I haven't taken calculas yet.
 
LSCupcake said:
I can't use f'(x) because we learn that in calculas, that is a derivative right? And I haven't taken calculas yet.

Yes, that's a derivative. If you can't use that then you must have been told something about minima of cubics that will let you get the third equation.
 
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