BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 3.1, Problem 79E
To determine

To find: The rough graph of the parabola of the function f(x)=(xm)(xn) , x -intercepts of the graph of f , x -coordinate of the vertex of parabola in terms of m and n . Confirm the answers by expanding and with the help of formulas.

Expert Solution

Answer to Problem 79E

The graph of the parabola f(x)=(xm)(xn) is given in Figure(1) when m=1 and n=2 . 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 .

Explanation of Solution

Given information:

The function is f(x)=(xm)(xn) .

Calculation:

Take a particular case when m=1 and n=2 . Sketch the graph of the function f(x)=(xm)(xn) .

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 3.1, Problem 79E

Figure(1)

By investigation of the function f(x)=(xm)(xn) with the help of graphing calculator, following observations are noted:

    Function y=(xm)(xn)mnx -intercepts of the graph y -intercept of graph Axis of symmetry
    y=(x2)(x8)282 and 816x=5
    y=(x2)(x3)232 and 36x=2.5
    y=(x3)(x5)353 and 515x=4
    y=(x+3)(x+5)353 and 515x=4
    y=(x+2)(x+4)242 and 416x=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)=(xm)(xn)=x2mxnx+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=(xm+n2)2+mnm2+n2+2mn4=(xm+n2)2+4mnm2n22mn4

Further simplify,

  f(x)=(xm+n2)2+4mnm2n22mn4=(xm+n2)2+m2n2+2mn4=(xm+n2)2m2+n22mn4=(xm+n2)2(mn)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=(xm+n2)2(mn)24(xm+n2)2=(mn)24xm+n2=±mn2x=±mn2+m+n2

Further simplify,

  x=±mn2+m+n2x=mn2+m+n2,(mn)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 .

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