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
$y-y1=m\left(x-x1\right)^2$y−y1=m(x−x1)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(x−2)2+9 or alternatively $f\left(x\right)-9=3\left(x-2\right)^2$ƒ (x)−9=3(x−2)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)−(9−5)=3(x−(2+2))2
or
$g\left(x\right)-4=3\left(x-4\right)^2$g(x)−4=3(x−4)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(x−4)2 cannot yeild a value >0 and as a result, $3\left(x-4\right)^2+4$3(x−4)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(x−4)2
Attached Graph: