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2 years ago

1.

Mathematics

2 answers:

Airida [17]2 years ago

7 0

Answer:

C. y = |x| - 1

Step-by-step explanation:

y = |x| Is the same graph, but with the vertex on point (0,0)

y = |x+1| Moves graph to the left 1

y = |x| + 1 Moves the graph up 1

y = |x| - 1 moves the graph down 1

y = |x-1| moves the graph right one

Therefore, y = |x| - 1 is the answer.

To further explain it, you can write down a table of values for proof. For y = |x| when x = 0, y = 0. When x = 1, y = 1 and so on. When x = -1, y = 1 and so on in that direction.

If we look at the choices and plug in numbers, we get for A when x = 0, y = 1, which if we look at the graph given does not fit.

For B, when x = 0, we get y = 1, which again does not work.

For C, when x = 0, y = -1, which is the answer we need, but to make sure, I am going to do D.

For D, when x = 0, y = 1. Therefore, C is the only possible answer.

adoni [48]2 years ago

3 0

Answer:

y = |x| - 1

Step-by-step explanation:

The difference between the parent function (y=|x|) and the graph, is that the graph is 1 unit down. y = |x| -1 has the parent function 1 unit down

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In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$