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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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
$$f(x)=a(x-x_0)^2+y_0$$
$$y=x^2-4x$$
$$x_0=-\frac b{2a}=2; y_0=f(2)=-4$$
Hint:
$x^2-4x=(x-2)^2-4$
So, shift the graph $4$ units down and $2$ units to the right.