How do you find the domain, find where f is increasing, vertex, intercepts, maximum of #f(x)=-(x-2)^2+1#?

1 Answer
Jul 17, 2017

Vertex#->(x,y)=(2,1)) larr" A maximum"#

#x_("intercepts")=1 and 3#
#y_("intercept")=-3#

increasing for #x<2#

Domain#->"input"-> {x: -oo<=x<=+oo}#

Explanation:

As the coefficient of #x^2# is -1 the graph is of form #nn#

Thus there is a determinable maximum.

Set #y=-(x-2)^2+1#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the vertex")#

The given function is in the form #y=a(x+b/(2a))^2+k# which is the vertex form (completing the square). Thus with a slight modification the vertex may read off it directly.

#x_("vertex")=(-1)xxb/(2a)->(-1)xx(-2)=+2#

#y_("vertex")=k=1#

#color(green)(" Vertex"->(x,y)=(2,1)) #
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine x-intercepts")#

These occur at #y=0# so we have:

#y=0=-(x-2)^2+1#

Add #(x+2)^2# to both sides

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

Square root both sides

#x-2 =+-1#

Add 2 to both sides

#color(green)(x=2+-1->x=1 and 3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine y-intercept")#

Set #x=0# giving

#y_("intercept")=-(0-2)^2+1#

#color(green)( y_("intercept")=-4+1=-3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine where the function is increasing")#

As the graph form is #nn# then #f(x)# is increasing to the left of the maximum at #(2,1)#. So increasing for #x<2#

It is neither increasing or decreasing at #x=2#