Kinley Richmond

2023-02-27

How to graph $y=|-\frac{1}{4}x-1|$?

sarabol2zsr

determining the location of y=0
$-\frac{1}{4}x-1=0$
then $x=-4$
when $x\le -4$
$|-\frac{1}{4}x-1|=-\frac{1}{4}x-1$
when $x>-4$
$|-\frac{1}{4}x-1|=\frac{1}{4}x+1$
both lines 'll be draw
$y=-\frac{1}{4}x-1$
finding y-intercept when x=0
$y=-1$ then the point to ubicate $\left(0,-1\right)$
when y=0
$x=-4$ the point is $\left(-4,0\right)$
for $y=\frac{1}{4}x+1$
y-intercept is when x=0
$y=1$ $\left(0,1\right)$
x-intercept is when y=0
$x=-4$ $\left(-4,0\right)$

Lekno2y5t

Everything that is a negative number is a positive number according to absolute value.
I simply draw a net. Then put instead of x a number. For example:
x=0 then y=1
if x=-4 the y=0
Now you can draw a line that goes through this two points. The other part will be mirrored just like
the right side. Let's see.
Let's put instead of x a number -8. Then y=1 just as for the x=0.
Additionally, it is possible to divide this function into two (a bit useless for this particular example).
${f}_{1}:y=-\frac{1}{4}x-1$ for $x\ge 0$
${f}_{2}:y=\frac{1}{4}x+1$ for $x<0$

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