[SOLVED] polynomial with two unknowns 1. The problem statement, all variables and given/known data The graph of f(x)= 3x^4 + 14x^3 + px^2 + qx + 24 has x-intercepts -4 and 2. Determine the function. 3. The attempt at a solution I could solve it if there were only one unknown but i don't know how to do it if there are two unknowns. What i did so far is plug -4 for x and 0 for the output, got an expression = 0 did the same thing for 2 since both equal 0, i set them equal to each other, simplified and got: 2p-q=13.3 Don't know what to do next.
Rather then setting them equal to each other. Solve for p with one of your x-intercepts then plug it in your other set with the other x-intercept.
What roco is getting at is that you can create a system of two simultaneous equations thus; [tex]f(-4) = 0[/tex] [tex]f(2) = 0[/tex]
yes i know that, that is what i initially did. So you get 2 equations, both equal to 0, both have q and p in them. But i don't know what to do next.
1) 0 = 16p - 4q - 104 2) 0 = 4p +2q + 184 then i divided both by 2: 1) 0 = 8p - 2q - 52 2) 0 = 2p + q + 92
Well either way, your answer isn't right b/c your answer should have ended up being in terms of a solution. I'll go step by step. Let me type this up.
If the OP hasn't met simultaneous equations it's pretty safe to say that they haven't been introduced to linear algebra (which IMHO is over-kill for a question such as this). Firstly, what you have written is an expression, not an equation. Secondly, you might want to check your coefficent of p.
0 = 12p + 132 is an equation ya i typed it wrong checked the back of the book, got the right answer (f(x)= 3x^4 + 14x^3 - 11x^2 - 70x + 24). ok i know how to do them now, thnx hootenanny, funny name
sure got a quick question: is the graph of y = log3^(x+4) the same as the graph of y = log3^x+4? *the base is 3 not 10 I think the second one is y=log3^x moved up by 4 units.
This will only apply to linear equations. [tex]a_1 x+b_1 y=c_1[/tex] [tex]a_2 x+b_2 y=c_2[/tex] [tex]x=\frac{\left|\begin{array}{cc}c_1 & b_1 \\ c_2 & b_2\end{array}\right|}{\left|\begin{array}{cc}a_1 & b_1 \\ a_2 & b_2\end{array}\right|}=\frac{c_1 b_2 - b_1 c_2}{a_1 b_2 - b_1 a_2}[/tex] [tex]y=\frac{\left|\begin{array}{cc}a_1 & c_1 \\ a_2 & c_2\end{array}\right|}{\left|\begin{array}{cc}a_1 & b_1 \\ a_2 & b_2\end{array}\right|}=\frac{a_1 c_2 - c_1 a_2}{a_1 b_2 - b_1 a_2}[/tex] Notice that the denominator for both x & y are the same (determinant). While the numerators vary. You can also solve for 3 unknowns but that takes too long to type.