Write in function?

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2 Answers
Tony B
Feb 2, 2018

To get my graphing package to show the valid points on the graph I used inequalities. So it is the blue line above the green area.

Explanation:

I suspect they are looking for you to calculate the 'critical point' which in the case is the y-intercept. This is at #x=0# and sketch an approximation of the shape to the right of this point.

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

#y= | -[ (0+2)^2] + 1 |#

#y=| -4+1|#

#y=| -3 | = +3#

#y_("interecpt")->(x,y)=(0,3)#

Tony B

Douglas K.
Feb 2, 2018

Given: #f(x) = |-(x+2)^2+1|,0 <= x <2#

Expand the expression inside of the absolute value:

#f(x) = |-(x^2+4x+4)+1|,0 <= x <2#

Distribute the -1:

#f(x) = |-x^2-4x-4+1|,0 <= x <2#

Combine like terms

#f(x) = |-x^2-4x-3|,0 <= x <2#

Find the zeros of the quadratic:

#-x^2-4x-3=0#

#(x+1)(x+3)=0#

#x = -1 and x = -3#

Because the quadratic represents a parabola that opens downward, it is greater than or equal to zero within the domain, #-3 <=x <=-1#

This means that the absolute value function does nothing to the quadratic within this domain:

#f(x)= -x^2-4x-3, -3 <=x <=-1#

Outside of this domain, the absolute value function multiplies the quadratic by -1:

#f(x)= { (x^2+4x+3, x < -3), (-x^2-4x-3, -3 <=x <=-1), (x^2+4x+3, x > -1) :}#

The above is the piecewise functional description of #f(x)#

The interval [0,2) is included in the last piece:

#f(x) = x^2+4x+3, 0 <= x < 2#

Here is a graph of this:

www.desmos.com/calculator

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