How do you graph $y + 4 = 0$ $y+4=0$ ?

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Answer 1 · 2017-10-18T00:41:14

See a solution process below:

We can solve this for $y$ $y$ as:

$y + 4 - 4 = 0 - 4$ $y + 4 - color(red)(4) = 0 - color(red)(4)$

$y + 0 = - 4$ $y + 0 = -4$

$y = - 4$ $y = -4$

This equation shows for each and every value of $x$ $x$ then $y = - 4$ $y = -4$ .

We can find two points to plot as:

For $x = 0$ $x = 0$ then $y = - 4$ $y = -4$ or $(0, - 4)$ $(0, -4)$

For $x = 5$ $x = 5$ then $y = - 4$ $y = -4$ or $(5, - 4)$ $(5, -4)$

graph{(x^2+(y+4)^2-0.075)((x-5)^2+(y+4)^2-0.075)=0 [-15, 15, -7.5, 7.5]}

We can now draw a line through the two points giving:

graph{(y+4)(x^2+(y+4)^2-0.075)((x-5)^2+(y+4)^2-0.075)=0 [-15, 15, -7.5, 7.5]}

We can this is a horizontal line from $- 4$ $-4$ on the y-axis

Answer 2 · 2017-10-18T00:49:07

Refer to the explanation.

Graph:

$y + 4 = 0$ $y+4=0$

Solve for $y$ $y$ by subtracting $4$ $4$ from both sides of the equation.

$y = - 4$ $y=-4$

The graph will be a horizontal line at $y = - 4$ $y=-4$ because all of the points will be $(x, - 4)$ $(x, -4)$ , where $x = \pm \infty$ $x=+-oo$ .

How do you graph y+4=0y+4=0?