Question #92542

1 Answer
George C.
May 14, 2016

See explanation...

Explanation:

Note that vertex form of a vertical parabola can be written:

#y = a(x-h)^2+k#

where #a# is a multiplier determining the 'steepness' of the parabola and #(h, k)# is the vertex (turning point).

We are given:

#f(x) = 1-x#

#g(x) = 2x^2-9#

#h(x) = (g @ f)(x)#

#= g(f(x))#

#= g(1-x)#

#= 2(1-x)^2-9#

#= 2(1-2x+x^2)-9#

#= 2-4x+2x^2-9#

#= color(blue)(2x^2-4x-7)#

#= 2(x^2-4x+4)-15#

#= 2(x-2)^2-15#

#= color(blue)(2(x+(-2))^2+(-15))#

The minimum value of #h(x)# occurs when #(x-2)^2 = 0#. That is when #x=2#. hence the turning point is at #color(blue)(""(2, -15))#

If #k(x)# is a translation of #h(x)# then it must have the same multiplier #2#. Given that its turning point is at #(-1, -5)#, its equation can be written:

#color(blue)(k(x) = 2(x+1)^2-5)#

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