Function

**Ellla** · December 8th 2010, 08:37 AM

Could anyone help me on this:

how to prove, that with all kind of a, b and c value, function's $f(x) = (x-a)(x-b)-c^2$ graph and Ox axis has at least one common point.

What should I start at? Could you lead me to the right direction? thank you.

**Soroban** · December 8th 2010, 09:25 AM

Hello, Ellla!

$\text{Prove that, for all values of }a,b,c,\text{ the graph of the function:}$

. . $f(x) \:=\: (x-a)(x-b)-c^2\,\text{ has at least one }x\text{-intercept.}$

An $\,x$ -intercept occurs when $f(x) = 0.$

We have: . $(x-a)(x-b) - c^2 \;=\;0$

. . . . $x^2 - (a+b)x + (ab-c^2) \:=\:0$

Quadratic Formula:

. . $x \;=\;\dfrac{(a+b) \pm\sqrt{(a+b)^2 - 4(ab-c^2)}}{2}$

. . . . $=\;\dfrac{(a+b) \pm\sqrt{a^2 + 2ab + b^2 - 4ab + 4c^2}}{2}$

. . . . $=\;\dfrac{(a+b)\pm\sqrt{a^2 - 2ab + b^2 + 4c^2}}{2}$

. . . . $=\;\dfrac{(a+b) \pm\sqrt{(a-b)^2 + 4c^2}}{2}$ .[1]

If $a = b = c = 0$ , [1] has one value.
. . The function has one $\,x$ -intercept.

For any other values of $a,b,c$ , [1] has two values.
. . The function has two $x$ -intercepts.

**emakarov** · December 8th 2010, 09:26 AM

One way it to see that f(a) = -c^2 <= 0. On the other hand, if x0 > max(|a|, |b|) + |c|, then f(x0) > 0. Therefore, f(x) must assume all intermediate values between f(a) and f(x0), including 0.

Strictly speaking, this proof invokes the Intermediate value theorem from calculus, but it is one of the most obvious-looking theorems (not to say that its proof is trivial). Students of calculus are often puzzled why it has to be proved at all.

**Ellla** · December 8th 2010, 11:46 AM

Thank you.

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