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The graph of the function g(x) has the same shape and direction of opening as the graph of f(x) = 3 (x-2)^2 + 9. The graph of g(x) has a vertex that is 2 units to the right and 5 units down from the vertex of the graph of f(x). 

a. Determine an equation of the function g(x).

 

 

b. State the domain and range of g(x). 

Domain: 

Range:

 

c. Write another function h(x) with the same vertex and shape, but whose graph opens in the opposite direction. 

6 years ago

Answered By Kevin G

a) For equations in the form

   $y-y1=m\left(x-x1\right)^2$yy1=m(xx1)2 ,

the location of the vertex is at the point (x1,y1)

Therefore, the vertex of the graph of  

   $f\left(x\right)=3\left(x-2\right)^2+9$ƒ (x)=3(x2)2+9   or alternatively  $f\left(x\right)-9=3\left(x-2\right)^2$ƒ (x)9=3(x2)2 

is at the point (2,9).

To move the vertex 2 units to the right and 5 units down, we need to add 2 to the value of x1 and subtract 5 from x2 giving us

    $g\left(x\right)-\left(9-5\right)=3\left(x-\left(2+2\right)\right)^2$g(x)(95)=3(x(2+2))2   

or 

  $g\left(x\right)-4=3\left(x-4\right)^2$g(x)4=3(x4)2    

b) The domain of g(x) is $(-\infty,\infty)$(∞,∞) because you can put any real number in for x and the function will give you a real number back.

The range of g(x) is  $[4,\infty)$[4,∞) since  $3\left(x-4\right)^2$3(x4)2 cannot yeild a value >0 and as a result,  $3\left(x-4\right)^2+4$3(x4)2+4 cannot yeild a value  >4

c) Just switch the sign of the value of 'm' (which in this case is 3):

 $h\left(x\right)-4=-3\left(x-4\right)^2$h(x)4=3(x4)2 

 

Attached Graph: