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I know about transformations and how to graph a function like $f(x) = x^2 - 2$. We just shift the graph 2 units down. But in this case, there's an $-4x$ in which the $x$ complicated everything for me. I understand that the graph will be a parabola for the degree of the function is 2, but I'm not exactly sure how I can graph it.

I can take various values for $x$ and then calculate $f(x)$, but I don't wanna do that. So, how do I plot something like this?

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Hint:

$x^2-4x=(x-2)^2-4$

So, shift the graph $4$ units down and $2$ units to the right.

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  • $\begingroup$ so, just complete the square and then plot? $\endgroup$ – MathEnthusiast May 17 '16 at 4:04
  • $\begingroup$ @user331377 Yes, can you plot it now? $\endgroup$ – Dragonemperor42 May 17 '16 at 4:05
  • $\begingroup$ Absolutely! Thank you! $\endgroup$ – MathEnthusiast May 17 '16 at 4:05
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You can also factor it to get $$y=x (x-4), $$ so the parabola crosses the $x$-axis at $x=0$ and $x=4$. The vertex of the parabola must lie halfway between the intercepts, so it is at $x=2$; when $x=2$, we have $y=2 (2-4)=-4$.

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In general: Let $f(x)=ax^2+bx+c$ $$ax^2+bx+c=a\left( x^2+\frac ba x+\frac ca\right)=a\left( x^2+2x\cdot\frac b{2a}+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac ca\right)=$$ $$=a\left(\left(x+\frac b{2a}\right)^2+\frac{4ac-b^2}{4a^2}\right)=a\left(x+\frac b{2a}\right)^2+\frac{4ac-b^2}{4a}$$ Let $x_0=-\frac b{2a}; y_0=\frac{4ac-b^2}{4a}$. Then $$f(x)=a(x-x_0)^2+y_0$$ $$y=x^2-4x$$ $$x_0=-\frac b{2a}=2; y_0=f(2)=-4$$ enter image description here

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  • $\begingroup$ Can you take this to arbitrary polynomials? $\endgroup$ – Jacob Wakem Jul 3 '16 at 13:59

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