Finding X and Y Intercepts w/ Absolute Value: y = -|x + 5|

[SCSmith]

y = -|x + 5|

Let x = 0

y = -|0 + 5|
y = -|5|
y = -5

so the y- intercept is (0,-5)

Let y = 0

0 = -|x + 5|
0 = - (x + 5)
0 = -x - 5 add 5
5 = +or- x

so the x- intercept is (+or-5,0)

 

[arthur ohlsten]

example
y=lxl is a V shaped curve ,open up, vertex at 0,0
y=lx-al is a V shaped curve, open up, vertex at a,0
y=-lx-bl is a V shaped curve, open down, vertex at b,0

y=-lx+5l a V shaped curve , open down, vertex at -5,0
two straight lines y=x-5] for x<5
and y=-x+5 x>5

Arthur

 

[soroban]

Re: Finding X and Y Intercepts with Absolute Value.

Hello, SCSmith!

Your x-intercepts are incorrect . . .

Find the intercepts of: \(\displaystyle \:y \:=\:-|x\,+\,5|\)

Let \(\displaystyle x\,=\,0:\;\;y \:=\:-|0\,+\,5| \:=\:-|5| \:=\:-5\)

so the y- intercept is \(\displaystyle (0,-5)\;\) . . . Yes!

Let \(\displaystyle y\,=\,0:\;\;-|x\,+\,5|\:=\:0\;\;\Rightarrow\;\;|x\,+\,5|\:=\:0\)

Then: \(\displaystyle \:x\,+\,5\:=\:\pm0\;\;\Rightarrow\;\;x\,=\,-5\)

There is one x-intercept: \(\displaystyle \,(-5,0)\)

 

[arthur ohlsten]

saroban is correct
I never really answered you
only one intercept [-5,0]

y=-lx+5l is a V shaped curve, open down, vertex at -5,0

two straight lines
y=-lx+5l can be replaced with :

y=-[x+5] x>-5
y=[x+5] x<-5

Arthur