Question

Need help on this Picture below Thanks .

Answer 1

a) (-1, 3)

b) Positive (up)

c) (-2, 4), (-1, 3), (0, 4), (1, 5)

Explanation:

Given absolute value equation:

`y=|x+1|+3`

Part a

The vertex form of an absolute value equation is:

`\large\boxedy = a`

where:

• (h, k) is the vertex.
• a is the leading coefficient.

In this case:

• a = 1
• h = -1
• k = 3

Therefore, the vertex of the given equation is (-1, 3).

`\hrulefill`

Part b

As the leading coefficient (a) is positive, the graph will be positive and open upwards.

If the leading coefficient was negative, the graph would open downwards.

`\hrulefill`

Part c

To find points on the graph, substitute the given x-values into the equation and calculate the corresponding y-values:

`\begin{aligned}x=-2 \implies y&=|-2+1|+3\\y&=|-1|+3\\y&=1+3\\y&=4\end{aligned}`

`\begin{aligned}x=-1 \implies y&=|-1+1|+3\\y&=|0|+3\\y&=0+3\\y&=3\end{aligned}`

`\begin{aligned}x=0 \implies y&=|0+1|+3\\y&=|1|+3\\y&=1+3\\y&=4\end{aligned}`

`\begin{aligned}x=1 \implies y&=|1+1|+3\\y&=|2|+3\\y&=2+3\\y&=5\end{aligned}`

Therefore, the completed table is:

`\begin{array}c\cline{1-2}\vphantom{\frac12}x&y\\\cline{1-2}\vphantom{\frac12}-2&4\\\cline{1-2}\vphantom{\frac12}-1&3\\\cline{1-2}\vphantom{\frac12}0&4\\\cline{1-2}\vphantom{\frac12}1&5\\\cline{1-2}\end{array}`

To sketch the graph:

• Plot the vertex at (-1, 3).
• Plot the other points from the table.
• Draw a straight line from the vertex through the points to the right of the vertex.
• Draw a straight line from the vertex through the points to the left of the vertex.