# Ask Uncle Colin: A Curve

Dear Uncle Colin,

I’m given that a curve has equation $y = ax^3 + bx^2 + cx + 1$. It has a turning point at $\left( -1, \frac{11}{3} \right)$ and an inflexion point when $x=2$. How do I find the missing constants?

- I’m Not Feeling Like Evaluating Constants, Thanks

Hi, INFLECT, and thanks for your message!

To find these constants, you need to use the information you’re given - and there are three pieces here that look like two.

The first is that the curve passes through $\left(-1, \frac{11}{3}\right)$.

The second is that the first derivative is zero there.

And the third is that the second derivative is zero when $x=2$.

## So let’s maths those up

When $x=-1$, $y = -a + b - c + 1 = \frac{11}{3}$.

That gives $-3a + 3b - 3c = 8$.

Differentiating, we get $\dydx = 3ax^2 + 2bx + c$, and when $x=-1$, that gives $3a- 2b + c = 0$

And differentiating again, $\diffn{2}{y}{x} = 6ax + 2b$; when $x=2$, that gives us $12a + 2b = 0$ - or $b = -6a$.

## Substituting back

In our second equation, we now have $15a + c = 0$, so $c = -15a$.

Putting both things into the first equation gives $-3a - 18a + 45a = 8$, so $a = \frac{1}{3}$.

That leads directly to $b = -2$ and $c = -5$.

So your equation is $y = \frac{1}{3}x^3 - 2x^2 - 5x + 1$.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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##### Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.