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euclid2 · #5 December 3rd, 2008, 03:29 PM

Quote:

Originally Posted by Soroban

Hello, euclid2!

There are several approaches to this problem.
. . Here's one of them . . .

If the quadratic, $ax^2 + bx + c$ , has roots $p\text{ and }q$ , then: . $p+q \:=\:-\frac{b}{a},\;\;pq \:=\:\frac{c}{a}$

So we have: . $\begin{array}{ccc}p+q &=&4 \\ pq &=& 1\end{array}\quad\Rightarrow\quad\begin{array}{ccc}\frac{b}{a} &=& \text{-}4 \\ \frac{c}{a} &=& 1 \end{array}$ . $\Rightarrow\quad\begin{array}{ccc}b &=&\text{-}4a \\ c &=&a \end{array}$

. . The function (so far) is: . $f(x) \:=\:ax^2 -4ax + a$

Since (1,-2) is on the graph: . $a\!\cdot\!1^2 - 4a\!\cdot\!1 + a \:=\:-2 \quad\Rightarrow\quad -2a \:=\:-2$

. . Hence: . $a\:=\:1,\;\;b\:=\:-4,\;\;c\:=\:1$

Therefore: . $f(x)\;=\;x^2 - 4x + 1$

Thank you very much, although can you edit this to show me where you are getting those numbers from? Thanks, again.