How do you graph #f(x)=3(x-4)^2+2# and identify the vertex, axis of symmetry, domain, range, max or min values, increasing and decreasing intervals?

1 Answer
Narad T.
Oct 31, 2016

The vertex is #(4,2)#
The axis of symmetry is #x=4#
The domain is #RR#
The range is #(4<=f(x)<+oo)#
The min value is #(4,2)#

Explanation:

The vertex is when #x=4##=>##f(x)=3*0+2=2#
So the vertex is at #(4,2)#
The axis of symmetry is #x=4#
For a polynomial function, the domain is #RR#
We need to calculate the derivative
#f'(x)=6(x-4)#
We have min or max when #f'(x)=0#
So, #6(x-4)=0# #=># #x=4#
We can do a chart
#x##color(white)(aaaaa)##-oo##color(white)(aaaaa)##4##color(white)(aaaaa)##+oo#
#f'(x)##color(white)(aaaaa)##-##color(white)(aaaaa)##+#
#f(x)##color(white)(aaaaaa)##darr##color(white)(aaaaa)##uarr#

So we have a min at #x=4#
So the range is #(4<=f(x)<+oo)#
The decreasing values are # -oo < x < =4 #
The increasing values are #4<=x<+oo#

graph{3(x-4)^2+2 [-9.94, 12.56, -2.92, 8.33]}

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