Given:
[tex]f(x)=x^2+2x-6[/tex]
We get two points from the given table.
(-6,14) and (-3,8).
Required:
We need to find the range of (f+g)(x).
Explanation:
Consider the equation of the linear function.
[tex]g(x)=mx+b[/tex]
where m is the slope.
Consider the formula to find the slope.
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex][tex]Substitute\text{ }y_2=8,y_1=14,x_2=-3,\text{ and }x_1=-6\text{ in the formula to find the slope m.}[/tex]
[tex]m=\frac{8-14}{-3-(-6)}[/tex]
[tex]m=\frac{-6}{3}=-2[/tex]
[tex]Substitute\text{ m=-3 in the equation }g(x)=mx+b.[/tex][tex]g(x)=-2x+b[/tex][tex]Substitute\text{ x=-6 and g\lparen-6\rparen=14 in the equation to find the value of b.}[/tex]
[tex]14=-2(-6)+b[/tex]
[tex]14=12+b[/tex]
Subtract 12 from both sides of the equation.
[tex]14-12=12-12+b[/tex]
[tex]2=b[/tex]
[tex]Substitute\text{ b =2 in the equation }g(x)=-2x+b.[/tex][tex]g(x)=-2x+2[/tex]
[tex]We\text{ know that }(f+g)(x)=f(x)+g(x).[/tex][tex]Substitute\text{ }f(x)=x^2+2x-6\text{ and }g(x)=-2x+2\text{ in the equation.}[/tex]
[tex](f+g)(x)=(x^2+2x-6)+(-2x+2)[/tex]
[tex]=x^2+2x-6-2x+2[/tex]
[tex]=x^2+2x-2x-6+2[/tex]
[tex]=x^2-4[/tex][tex]\text{We get }(f+g)(x)=x^2-4.[/tex]
The graph of the function (f+g)(x).
We know that the range of a graph consists of all the output values shown on the y-axis.
The minimum value of the range is -4.
The graph moves upward to infinity.
The maximum value of the range is infinity.
[tex]range=(-4,\infty)[/tex]
Final answer:
[tex]range=(-4,\infty)[/tex]
