Write $2+0.8x-0.04x^2$ in the form $A-B(x+C)^2$, where A, B and C are constants to be determined.
Here's how I have tried it out:
$2+0.8x-0.04x^2$
$-0.04x^2+0.8x+2$
$-0.04[(x-10)^2-10]+2$
$-0.04(x-10)^2+o.8+2$
$2.8-0.04(x-10)^2$
So the answer should be, $A=2.8$ $B=0.04$ $C=-10$. But on my book solution sheet, $A = 6$. Could you please help me out what have I done wrong?
$\endgroup$3 Answers
$\begingroup$It should be
$$-0.04[(x-10)^2 - 10^{\color{red}2}]+2=-0.04(x-10)^2+\color{blue}4+2$$
$\endgroup$ 1 $\begingroup$I could just point out the error, but a user already did that. So, let me rather give you two ways of trying to find an error.
Method 1: "try it out"
In this method, you pick a value of $x$, and try to plug it into the beginning and then each line of the derivation. You'll want to pick "simple" values of $x$ to try and find the error. In particular, let's take $x=0$. Your first line becomes $2\cdot 0.8\cdot 0 - 0.04\cdot 0^2 = 2$, but take a look at the third line:
Take $-0.04[(x-10)^2-10]+2$, now plug in $x=0$, and you get$$-0.04[(0-10)^2-10]+2 = -0.04\cdot(10^2-10) + 2 = -3.6+2=-1.6\neq 2$$
so clearly, this is the line where something went wrong.
Method two: "work backwards"
In this method, take each line, and try to derive from it the line before. In particular, taking again the second line, you would get
$$\begin{align}-0.04[(x-10)^2-10]+2 &= -0.04((x^2-20x+100)-10)+2 \\&= -0.04\cdot(x^2-20x+90)+2 \\&= -0.04x^2+0.8x-3.6+2\\&=-0.04x^2+0.8x-1.6\end{align}$$
which is clearly not correct.
$\endgroup$ $\begingroup$Let
$$2+0.8x-0.04x^2=-\frac4{100}\left(x^2-20x-50\right)=-\frac4{100}\left(x^2-20x+100-150\right)$$
$$-\frac4{100}\left((x-10)^2-150\right)=-0.04(x^2-10)^2+6$$
$\endgroup$
