Analyze and sketch the graph of $f(x) = x^4 - 12x^3 + 48x^2 - 64x$

I'm currently doing AP Calculus AB homework, and need some help with this problem. It sucks to forget. I know how to get the first and second derivative. I know that critical numbers can be found using the first derivative, and that points of inflection and concavity intervals can be found by using the second derivative, but I honestly forgot how to do it.

Analyze and sketch the graph of $f(x) = x^4 - 12x^3 + 48x^2 - 64x$

Edit: Solved!

• First Derivative: $4x^3 - 36x^2 + 96x - 64$
• Second Derivative: $12x^2 - 72x + 96$
• X-Intercepts: $(0,0)$ $(4,0)$
• Y-Intercepts: $(0,0)$
• Critical Numbers: $X = 1$, $X = 4$
• Points of Inflection: $X = 2$, $X = 4$
• Increasing Intervals: $(1, Infinity)$
• Decreasing Intervals: $(-Infinity, 1)$
• Concavity Intervals: $(-Infinity, 2)$ $(2,4)$ $(4, Infinity)$
• Relative Minimums: $X=1$
• Relative Maximums: $X=2$

This should bring you forward:

• X- Intercepts: f(x) =0
• Y- Intercepts: f(0)
• critical numbers: f'(x) = 0
• Points of inflections: f''(x) = 0 and f'''(x) $\neq$ 0
• Increasing intervals: f'(x) > 0
• Decreasing intervals: f'(x) < 0
• Concavity upward: f''(x) > 0
• Concavity downward f''(x) < 0
• Relative Minimums: f'(x) = 0 (see above: critical numbers) and f''(x)>0
• Relative Maximums: f'(x) = 0 (see above: critical numbers) and f''(x)<0

However, you should - at least once - really understand those criteria (e.g. graphically) rather than just apply them.

• I'm not really too sure on how to find the second X-intercept (one of them being 0), the Y-intercept is definitely 0 due to every number having an X variable. I'm confused when it comes to setting the first derivative equal to 0, I've factored out 4 and now i'm left with $4(x^3 - 9x^2 + 14x - 16) = 0$ – Khon Duong Jan 9 '14 at 10:15
• For the second X-intercept: factor out x and then try out the divisors of the number without x (which will be 64). So one of the number 1,2,4,8,16,32,64 or their negative values should do it. Start with the smallest ones. Are you not allowed to use a calculator? – Bernd Jan 9 '14 at 10:27
• Not for this particular question, no. The main focus of this problem is to be able to completely or accurately sketch a graph based on the characteristics that you get from doing all of those. – Khon Duong Jan 9 '14 at 10:35
• For the first derivative: You factored out 4. That's correct. However it's 24, not 14. So for a product to be zero, at least one factor has to be zero. That means your 2nd factor must be 0 (since 4 is not). Again here: try divisors of the absolute number (without x: 16). Start with the smallest: 1,2, .. – Bernd Jan 9 '14 at 10:37
• Just for clarification, we're solving for the critical number right? And the problem is still $4(x^3 - 9x^2 + 14x -16)$ and you want me to factor it a second time? – Khon Duong Jan 9 '14 at 10:47