Using Table to Graph Lines Exs

0 3 since this is not a true statement there is no

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Unformatted text preview: fore, another point that sa+sfies the equa+on is (6, 3). Let us plug in x =  ­6 into the equa+on. y = (3/4)( ­6) – 3/2 =  ­9/2 – 3/2 =  ­12/2 =  ­6 Graph of  ­3x + 4y =  ­6 Now we must find at least two addi+onal points. Let us plug in x = 6 into the equa+on. y = (3/4)(6) – 3/2 = 9/2 – 3/2 = 6/2 = 3 Therefore, another point that sa+sfies the equa+on is (6, 3). Let us plug in x =  ­6 into the equa+on. y = (3/4)( ­6) – 3/2 =  ­9/2 – 3/2 =  ­12/2 =  ­6 Therefore, another point that sa+sfies the equa+on is ( ­6,  ­6). Graph of  ­3x + 4y =  ­6 Now we must find at least two addi+onal points. Let us plug in x = 6 into the equa+on. y = (3/4)(6) – 3/2 = 9/2 – 3/2 = 6/2 = 3 Therefore, another point that sa+sfies the equa+on is (6, 3). Let us plug in x =  ­6 into the equa+on. y = (3/4)( ­6) – 3/2 =  ­9/2 – 3/2 =  ­12/2 =  ­6 Therefore, another point that sa+sfies the equa+on is ( ­6,  ­6). We will plot the following points in the table, and the connect the points together with a straight line: X y  ­6  ­6 0  ­3/2 2 0 6 3 Graph of  ­3x + 4y =  ­6 It is easy to see by looking at the graph that this equation is a function and that its domain and range are all real numbers. Linear Equa+ons, y = a Example. Graph y = 3. Linear Equa+ons, y = a Example. Graph y = 3. Here, no ma\er what the value of x is, y is always 3. Linear Equa+ons, y = a Example. Graph y = 3. Here, no ma\er what the value of x is, y is always 3. We will find the x ­intercept by plugging in y = 0 into the equa+on and solving for x. Linear Equa+ons, y = a Example. Graph y = 3. Here, no ma\er what...
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