Given information:
The function is f(x)=(x−m)(x−n) .
Calculation:
Take a particular case when m=1 and n=2 . Sketch the graph of the function f(x)=(x−m)(x−n) .
Figure(1)
By investigation of the function f(x)=(x−m)(x−n) with the help of graphing calculator, following observations are noted:
Function y=(x−m)(x−n) | m | n | x -intercepts of the graph | y -intercept of graph | Axis of symmetry |
y=(x−2)(x−8) | 2 | 8 | 2 and 8 | 16 | x=5 |
y=(x−2)(x−3) | 2 | 3 | 2 and 3 | 6 | x=2.5 |
y=(x−3)(x−5) | 3 | 5 | 3 and 5 | 15 | x=4 |
y=(x+3)(x+5) | −3 | −5 | −3 and −5 | 15 | x=−4 |
y=(x+2)(x+4) | −2 | −4 | −2 and −4 | 16 | x=−3 |
So, from the observations it can be concluded that the x -intercepts of the graph of f are m and n .
From the table, it is visible that the x coordinate of the vertex of parabola is m+n2 .
Now check the results by expanding.
f(x)=(x−m)(x−n)=x2−mx−nx+mn=x2−(m+n)x+mn
Factorize the function.
f(x)=x2−(m+n)x+mn=x2−(m+n)x+mn+(m+n2)2−(m+n2)2=(x−m+n2)2+mn−m2+n2+2mn4=(x−m+n2)2+4mn−m2−n2−2mn4
Further simplify,
f(x)=(x−m+n2)2+4mn−m2−n2−2mn4=(x−m+n2)2+−m2−n2+2mn4=(x−m+n2)2−m2+n2−2mn4=(x−m+n2)2−(m−n)24
So, the x -coordinate of vertex of the parabola is m+n2 .
Now substitute f(x)=0 for the x -intercepts of parabola.
0=(x−m+n2)2−(m−n)24(x−m+n2)2=(m−n)24x−m+n2=±m−n2x=±m−n2+m+n2
Further simplify,
x=±m−n2+m+n2x=m−n2+m+n2,−(m−n)2+m+n2x=2m2,2n2x=m,n
So, the x -intercepts of parabola are m and n .
Therefore, the x -intercepts of the graph of f are m and n . The x -coordinate of the vertex of parabola in terms of m and n is equal to m+n2 .