How to determine $x$ and $y$ intercepts for $y = 4(x - 2)^2(x + 2)^3$

I need help to determine $x$ and $y$ intercepts for $$y = 4(x - 2)^2(x + 2)^3$$

I guess my first question is, do I need to get the equation into $$ax^3 + bx^2 + cx + d$$ form before starting?

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Would I then just use the x - 2 and x + 2 to determine that the x-intercepts are at 2 and -2? When I graph it, that shows that would be correct. – erimar77 Nov 8 '11 at 21:35
In general, the more "structure" an expression has, the easier it is to solve problems about it. Multiplying out tends to destroy structure. – André Nicolas Nov 8 '11 at 21:36

The $x$-intercepts are defined to be the points where $y=0$. Since what you have is factored, it is easy to see where $y=0$. If $y=0$ then one of the factors must be $0$. The $y$-intercept is where $x=0$. If it exists, it is unique. Just plug in $x=0$ and see what you get.

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Would I then just use the x - 2 and x + 2 to determine that the x-intercepts are at 2 and -2? – erimar77 Nov 8 '11 at 20:48
@erimar77: Yes. – Joe Johnson 126 Nov 8 '11 at 22:40

That would be counterproductive.

The $x$-intercepts are solutions to $y=0$, and your equation is set up so that this is easy to solve.

For the $y$-intercept, just set $x=0$ and calculate $y$.

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Multiplying is easy; factoring is hard. (Believe it or not, that's why it's hard for identity thieves to steal your money in properly secured e-commerce transactions. Unless maybe the thief is the Prince of Mongolia and you're inexperienced.)

If you had a polynomial in the form $ax^3 + bx^2 + cx +d$, you'd have to factor it to find the $x$-intercepts. This one you've already got in factored form.

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The x-intercepts can be calculated like this:

$$y = 4(x - 2)^2(x + 2)^3$$ $$y = 0$$

$$a \cdot b \cdot c = 0 \rightarrow a = 0 \vee b=0 \vee c=0$$

$$4 = 0 \vee (x-2)^2 = 0 \vee (x+2)^3 = 0$$

$$None \vee x = 2 \vee x \ -2$$

The y-intercept can be found by substituting $0$ for $x$ in $4(x - 2)^2(x + 2)^3$.

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The zeroes of the equation $y = 4(x-2)^2(x+2)^3$ are clearly defined as $x = +2, -2$ which everyone stated. Another idea which I think is interesting is the equations behaviour around these roots. The factor $(x-2)^2$ has an order of $2$ therefore near this root the equation will "bounce" or act quadratically. The factor $(x+2)^3$ has an order of 3 therefore the equation will act cubically near this point, with an inflection at the root.

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