Use the function f(x) = x2 − 2x + 8 and the graph of g(x) to determine the difference between the maximum value of g(x) and the minimum value of f(x).


a parabola that opens down and passes through 0 comma 3, 3 comma 12, and 5 comma 8


2

5

7

12

Use the function fx x2 2x 8 and the graph of gx to determine the difference between the maximum value of gx and the minimum value of fxa parabola that opens dow class=

Respuesta :

Answer:  b) 5

Step-by-step explanation:

As given by the graph, the maximum value of g(x) is 12

Now, let's find the Axis Of Symmetry of f(x). Then plug in the x-value into f(x) to find the minimum value of f(x).

[tex]x=\dfrac{-b}{2a}=\dfrac{-(-2)}{2(1)}=\dfrac{2}{2}=1\\\\f(x)=x^2-2x+8\\f(1) = 1^2-2(1)+8\\.\qquad =1-2+8\\.\qquad=7[/tex]

Difference = max of g(x) - min of f(x)

                 =       12          -       7

                 =                  5

Answer:

The difference between the  maximum value of g(x) and the minimum value of f(x) is:

                                    5

Step-by-step explanation:

  • The function f(x) is given by:

[tex]f(x)=x^2-2x+8[/tex]

which in vertex form is given by:

[tex]f(x)=(x-1)^2+7[/tex]

Now, we know that:

[tex](x-1)^2\geq 0\\\\(x-1)^2+7\geq 0+7\\\\(x-1)^2+7\geq 7\\\\i.e.\\\\f(x)\geq 7[/tex]

Hence, the minimum value of the function f(x) is:  7

  • Also, the function g(x) is a parabola that opens down and passes through (0,3) , (3,12)  and (5,8) .

Let the equation for g(x) be:

[tex]g(x)=ax^2+bx+c[/tex]

Now with the help of the passing through three points we get:

when it passes through (0,3)

[tex]3=a\times 0+b\times 0+c\\\\i.e.\\\\c=3[/tex]

when it pass through (3,12)

i.e.

[tex]9a+3b+c=12\\\\i.e.\\\\9a+3b+3=12\\\\9a+3b=12-3\\\\9a+3b=9\\\\3a+b=3---------(2)[/tex]

Also it pass through (5,8)

i.e.

[tex]25a+5b+c=8\\\\25a+5b+3=8\\\\25a+5b=8-3\\\\25a+5b=5\\\\5a+b=1-------(2)[/tex]

On subtracting equation (2) from equation (1) we have:

[tex]a=-1[/tex]

and by putting the value of  a in equation (1) we have:

[tex]b=6[/tex]

Hence, we get:

[tex]g(x)=-x^2+6x+3\\\\g(x)=-(x-3)^2+12[/tex]

Now, we know that:

[tex](x-3)^2\geq 0\\\\-(x-3)^2\leq 0\\\\-(x-3)^2+12\leq 12[/tex]

This means that the maximum value of g(x) is: 12

The difference between the maximum value of g(x) and the minimum value of f(x) is:

         12-7=5

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