Kinley Richmond

2023-02-27

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

sarabol2zsr

Beginner2023-02-28Added 6 answers

determining the location of y=0

$-\frac{1}{4}x-1=0$

then $x=-4$

when $x\le -4$

$\left|-\frac{1}{4}x-1\right|=-\frac{1}{4}x-1$

when $x>-4$

$\left|-\frac{1}{4}x-1\right|=\frac{1}{4}x+1$

both lines 'll be draw

$y=-\frac{1}{4}x-1$

finding y-intercept when x=0

$y=-1$ then the point to ubicate $(0,-1)$

when y=0

$x=-4$ the point is $(-4,0)$

for $y=\frac{1}{4}x+1$

y-intercept is when x=0

$y=1$ $(0,1)$

x-intercept is when y=0

$x=-4$ $(-4,0)$

$-\frac{1}{4}x-1=0$

then $x=-4$

when $x\le -4$

$\left|-\frac{1}{4}x-1\right|=-\frac{1}{4}x-1$

when $x>-4$

$\left|-\frac{1}{4}x-1\right|=\frac{1}{4}x+1$

both lines 'll be draw

$y=-\frac{1}{4}x-1$

finding y-intercept when x=0

$y=-1$ then the point to ubicate $(0,-1)$

when y=0

$x=-4$ the point is $(-4,0)$

for $y=\frac{1}{4}x+1$

y-intercept is when x=0

$y=1$ $(0,1)$

x-intercept is when y=0

$x=-4$ $(-4,0)$

Lekno2y5t

Beginner2023-03-01Added 3 answers

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 $x<0$

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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