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1. **Graph y = 12x + 3 2. ** y = -3x + 4 y 9 T 구 8 16 5 다 2 1 19 18 454 - 6 3-2 1 6 a x 대 - 2 1 2 6 B 9 x 1 0 2 3 10 . -6. R bo 를

1. **Graph y = 12x + 3 2. ** y = -3x + 4 y 9 T 구 8 16 5 다 2 1 19 18 454 - 6 3-2 1 6 a-example-1
User Prashant Jajal
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1 Answer

17 votes
17 votes

Answer:

Graphing the points we have;

Above is the graph of the given equation showing the derived points.

Step-by-step explanation:

Given the equation;


y=|2x|+3

To plot the graph we need to calculate the corresponding values of x and y at each point.

Let us calculate the values of y for x = -4,-2,0,2, and 4;


\begin{gathered} y=|2x|+3 \\ at\text{ x=-4;} \\ y=|2*-4|+3 \\ y=8+3 \\ y=11 \\ (-4,11) \end{gathered}
\begin{gathered} at\text{ x=-2} \\ y=|2*-2|+3 \\ y=4+3 \\ y=7 \\ (-2,7) \end{gathered}
\begin{gathered} at\text{ x=0;} \\ y=|2*0|+3 \\ y=3 \\ (0,3) \end{gathered}
\begin{gathered} at\text{ x=2;} \\ y=|2*2|+3 \\ y=4+3 \\ y=7 \\ (2,7) \end{gathered}
\begin{gathered} at\text{ x=4;} \\ y=|2*4|+3 \\ y=8+3 \\ y=11 \\ (4,11) \end{gathered}

Therefore, Graphing the points we have;

Above is the graph of the given equation showing the derived points.

1. **Graph y = 12x + 3 2. ** y = -3x + 4 y 9 T 구 8 16 5 다 2 1 19 18 454 - 6 3-2 1 6 a-example-1
User Dgund
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2.5k points