How do you graph y=x^3-4x^2+x+6?

Jul 8, 2017

Factorise to find x-intercepts

Explanation:

Look at the constant at the end (6) and use the factors of that number to find one of your roots. For example, if you put x = 2, that gives you 8-16+2+6 which = 0 therefore (x-2) is a factor.

You now know that the equation is a combination of (x-2) and, because it is a cubic graph, a quadratic is left over.

$y = \left(x - 2\right) \left(a {x}^{2} + b x + c\right)$

Now as you look at the coefficients of each term, the coeff of ${x}^{3}$ is 1 therefore a must be 1

To find the coeff of ${x}^{2}$, it is -4 which must equal b - 2a (a combination of $b x$ x $x$ and $a x$ times $- 2$)

Therefore $b = - 2$

Finally, you need the c which is 6 and is $- 2 c$ from the expansion so $c = - 3$

Altogether this gives us $y = \left(x - 2\right) \left({x}^{2} - 2 c - 3\right)$

The quadratic can be factorised to give:

$y = \left(x - 3\right) \left(x + 1\right)$

to give $y = \left(x - 3\right) \left(x - 2\right) \left(x + 1\right)$

These three brackets give us our three x intercepts (-1, 2 and 3) and when x = 0, y = 6 which is the y intercept!!

Don't forget cubic graphs are like sideways 's'