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2 years ago

5

1. What is the graph of the absolute value equation? y=|x|-9

2 answers:

Kruka [31]2 years ago

8 0

The graph of y = |x| - 9 is V shaped with vertex at (0, -9).

The graph of y = |x + 4| is V shaped with vertex at (-4, 0).

The graph of y = |x + 3| - 4 is V shaped with vertex at (-3, -4).

The graph of y = -1/2|x| is V shaped with vertex at (0, 0)

Bond [772]2 years ago

4 0

Using a graphing tool

We're proceeding to graph each of the cases

<u>case 1)</u> $y=\left|x\right|-9$

<u>The graph in the attached figure N $1$ </u>

The vertex of the graph is the point $(0,-9)$

The x-intercepts are the points $(-9,0)$ and $(9,0)$

<u>case 2)</u> $y=\left|x+4\right|$

<u>The graph in the attached figure N $2$ </u>

The vertex of the graph is the point $(-4,0)$

The y-intercept is the point $(0,4)$

<u>case 3)</u> $y=\left|x+3\right|-4$

<u>The graph in the attached figure N $3$ </u>

The vertex of the graph is the point $(-3,-4)$

The y-intercept is the point $(0,-1)$

The x-intercepts are the points $(-7,0)$ and $(1,0)$

<u>case 4)</u> $y=-0.5\left|x\right|$

<u>The graph in the attached figure N $4$ </u>

The vertex of the graph is the point $(0,0)$

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Elanso [62]

Answer:

x=-2 (the actual solution)

x=-5 (the extraneous solution)

Step-by-step explanation:

First step is to isolate the radical part (the part with the square root).

So we will add 4 on both sides first:

$\sqrt{x+6}=x+4$

To get rid of square we are going to square both sides:

$x+6=(x+4)^2$

We are going to expand right hand side (you can use "F.O.I.L":

$x+6=x^2+4x+4x+16$

Subtract x on both sides and subtract 6 on both sides:

$0=x^2+4x+4x-x+16-6$

Combine like terms:

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We will now factor:

$0=(x+2)(x+5)$ (Since 2(5)=10 and 2+5=7)

Set both factors equal to 0:

x+2=0 or x+5=0

Solve first equation by subtracting 2 on both sides to obtain x=-2.

Solve second equation by subtracting 5 on both sides to obtain x=-5.

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$\sqrt{x+6}-4=x$ for $x=-2$ :

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$\sqrt{4}-4=-2$

$2-4=-2$

$-2=-2$ is a true equation so $x=-2$ is indeed a solution.

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$\sqrt{-5+6}-4=-5$

$\sqrt{1}-4=-5$

$1-4=-5$

$-3=-5$ is false so $x=-5$ is what we call the extraneous solution.

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