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.
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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.
Hello, sjenkins!
Make a sketch . . .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.
We can see that a parabola will work.Code:|
| (3½,5)
| *
|
|
|
- - + - * - - - * - -
| 2 5
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: .$\displaystyle f(x) \:=\;-\frac{20}{9}(x-2)(x-5) \quad\Rightarrow\quad\boxed{ f(x) \;=\;-\frac{20}{9}x^2 + \frac{140}{9}x - \frac{200}{9}}$
