# Semi-circle function

• Sep 1st 2008, 08:27 AM
sjenkins
Semi-circle function
The graph of y=(3x-x)^.5 is a semi-circle with diameter coincident with the positive x-axis, radius 1.5, center at (1.5,0). create a function passing through (3.5,5) and intersecting the x-axis at x=2 and x=5.
• Sep 1st 2008, 11:38 AM
Soroban
Hello, sjenkins!

Quote:

The graph of y=(3x-x)^.5 is a semi-circle with diameter coincident
with the positive x-axis, radius 1.5, center at (1.5,0).

. . All this is all true . . . but who needs it ??

Create a function passing through (3.5, 5) and intersecting the x-axis at x=2 and x=5.

Make a sketch . . .
Code:

        |         |    (3½,5)         |      *         |         |         |     - - + - * - - - * - -         |  2      5
We can see that a parabola will work.

From the two x-intercepts, we have: .$\displaystyle f(x) \:=\:a(x-2)(x-5)$

To pass through $\displaystyle \left(\frac{7}{2},\:5\right)$, we have: .$\displaystyle a\left(\frac{7}{2}-2\right)\left(\frac{7}{2} - 5\right) \:=\:5$

. . $\displaystyle a\left(\frac{3}{2}\right)\left(-\frac{3}{2}\right) \:=\:5 \quad\Rightarrow\quad -\frac{9}{4}\,a \:=\:5 \quad\Rightarrow\quad a \:=\:-\frac{20}{9}$

Therefore: .$\displaystyle f(x) \:=\;-\frac{20}{9}(x-2)(x-5) \quad\Rightarrow\quad\boxed{ f(x) \;=\;-\frac{20}{9}x^2 + \frac{140}{9}x - \frac{200}{9}}$