Solving a Quadratic Equation "Using a Table and a Graph"

Question

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?

Answer 1

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.

Answer 2

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?

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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 $x$?

Answer 3

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.