Question

How do you use end behavior, zeros, y intercepts to sketch the graph of `f(x)=x^3+11x^2+35x+32`?

Answer 1

See explanation

Explanation:

End behavior: the x^3 term is dominant, which means that as x goes out to increasing values of positive x, (or to increasingly NEGATIVE values of negative x), f(x) increases towards positive infinity (or negative infinity), respectively.

y intercept: set x = 0, then f(x) is trivially read off as 32.

Zeroes: Well, it's not trivially obvious where y = 0 would be found.
`f(x) = 0` when `(x^3 + 11x^2+35x) = -32`

Now, this question is under "precalculus", so what follows may be a bit above & beyond. But:

Take the first derivative, `f'(x) = 3x^2 + 22x + 35`.

The original function f(x) will change direction (from increasing to decreasing, or vice versa) when `f'(x) = 0`. It's a quadratic equation, with roots `-5` and `-2 1/3`

So you can mark points (x = -5, y = f(-5)) and (x = `-2 1/3`, y = `f(2 1/3)`)

or points (-5, 7) and (`-2 1/3`, `-2.47777`)

If you've had some calculus, you can take the second derivative of f(x) and do some more analysis, but this may not be totally necessary here. So, to graph this function:
- plot points (-5,7), (-2.3333, -2.47777) (call it 2.5) and (0, 32)
- draw a nice, swoopy curve coming up from negative infinity on the left, topping out at -5,7, then down to -2.333, -2.5, then up to (0, 32), and then on up to positive infinity on the right.

Or, since we live in 2017, you can let your browser graph it:

graph{x^3 + 11x^2 + 35x + 32 [-22.37, 17.63, -8.72, 11.28]}

GOOD LUCK