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I need to find $a\in \Bbb Z, 0\le a\lt10 : f(1 + \frac{a}{10}) = 0$ for a number of different quadratic functions, for example $f(x) = -x^2 + 4x - 3$, by "using a table and a graph". Can someone explain what that actually is, and how one uses the specified methods to solve this sort of problem?

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$$-(x-a)(x-1)=-x^2+4x-3$$ –  lab bhattacharjee Dec 20 '13 at 15:45
    
Could you explain that please –  Desiree Dec 20 '13 at 15:49
    
If $a,1$ be the zeros, see en.wikipedia.org/wiki/Quadratic_equation#Vieta.27s_formulas –  lab bhattacharjee Dec 20 '13 at 15:51
    
Here's an idea: why don't you pay attention in class next time? –  AJMansfield Dec 20 '13 at 16:12
1  
You don't think I am... I do online @AJMansfield. Its harder to grasp things. Im just trying to get help. You don't need to comment. –  Desiree Dec 20 '13 at 16:14

3 Answers 3

up vote 3 down vote accepted

It would help you to factor $$f(x) = -x^2 + 4x - 3 = -(x - 3)(x - 1)$$

Then $$f(x) = 0 \iff x - 3 = 0 \;\text{ or } \; x - 1 = 0$$

As suggested, did you graph the function? You can then visually see where $f(x)$ intersects the $x$-axis: those are the "zeros" of the function.

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Yes, I did graph it using an online calculator, but the parabola does not cross exactly at an x-axis. –  Desiree Dec 20 '13 at 15:51
    
so then I would solve x-a=0 and x-1=0.. a= -1, right? –  Desiree Dec 20 '13 at 15:54
    
Yes, it does: it crosses the x-axis when $f(x) = 0$; and $f(x) = 0$ if and only if $x = 3$ or $x = 1$. Per your question, yes, you solve each of $x - 3 = 0$, $x - 1 = 0$, separately. There are two values of $x$ that are zeros: one is $x = 3$, the other is $x = 1$. –  amWhy Dec 20 '13 at 15:54
    
Why would we use 3.. for the equation x-3=0 if the equation uses a and a would be the first number. I guess I don't understand on that part. –  Desiree Dec 20 '13 at 16:01
    
And my graph shows that the parabola is close to -5 and close to 1 but not exactly on it. –  Desiree Dec 20 '13 at 16:01

Here is how you use a table to solve the problem. Make a list of all the allowed values of $a$, the corresponding values of $x$, and the resulting value for $f(x)$

 a │  x  │ f(x)
───┼─────┼──────
 0 │ 1.0 │ ?.??
 1 │ 1.1 │ ?.??
 2 │ 1.2 │ ?.??
   ┊     ┊
 9 │ 1.9 │ ?.??

Now, looking at your table, for what value of $a$ does $f(x)$ come out as zero?


Here is how you use a graph to solve the problem. Plug the equation into a graphing utility, like this one. Now, looking at the graph, for what values of $x$ does $f(x)$ equal zero? For which of the two points can you pick an $a$ such that $1+\frac{a}{10}$ equals $x$?

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Okay so for the first part of the graph 0, 1.0, ?.?? I got 6 but that doesn't seem right. –  Desiree Dec 20 '13 at 16:40
    
@Desiree nope, try again. You are messing up order of operation. –  AJMansfield Dec 20 '13 at 20:19
    
@Desiree I just figured out what the error was: you switched the sign of the last term in the equation. You evaluated $−x^2+4x+3$, when the function is actually supposed to be $−x^2+4x-3$. –  AJMansfield Dec 21 '13 at 23:58

For the first one, try the quadratic formula (or factoring, but if you cannot see how to factor it, the quadratic formula will give you what you need anyway). You have that $x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ where the quadratic you have is of the form $y=ax^2+bx+c$. Therefore, in your case, $a=-1$, $b=4$, and $c=-3$. When you plug these into the formula, you get what you need.

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Thank you! :) it makes more sense now. –  Desiree Dec 20 '13 at 18:30

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