commongraphs

# commongraphs - xample 1 Graph Solution This is a line in...

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xample 1 Graph . Solution This is a line in the slope intercept form In this case the line has a y intercept of (0, b ) and a slope of m . Recall that slope can be thought of as Note that if the slope is negative we tend to think of the rise as a fall. The slope allows us to get a second point on the line. Once we have any point on the line and the slope we move right by run and up/down by rise depending on the sign. This will be a second point on the line. In this case we know (0,3) is a point on the line and the slope is . So starting at (0,3) we’ll move 5 to the right ( i.e. ) and down 2 ( i.e. ) to get (5,1) as a second point on the line. Once we’ve got two points on a line all we need to do is plot the two points and connect them with a line. Here’s the sketch for this line. Example 2 Graph Solution

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There really isn’t much to this problem outside of reminding ourselves of what absolute value is. Recall that the absolute value function is defined as, The graph is then, Example 3 Graph . Solution This is a parabola in the general form. In this form, the x -coordinate of the vertex (the highest or lowest point on the parabola) is and we get the y -coordinate is . So, for our parabola the coordinates of the vertex will be. So, the vertex for this parabola is (1,4).
We can also determine which direction the parabola opens from the sign of a . If a is positive the parabola opens up and if a is negative the parabola opens down. In our case the parabola opens down. Now, because the vertex is above the x -axis and the parabola opens down we know that we’ll have x -intercepts ( i.e. values of x for which we’ll have ) on this graph. So, we’ll solve the following. So, we will have x -intercepts at and . Notice that to make our life easier in the solution process we multiplied everything by -1 to get the coefficient of the positive. This made the factoring easier. Here’s a sketch of this parabola.

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## This note was uploaded on 11/06/2011 for the course MATH 151 taught by Professor Sc during the Fall '08 term at Rutgers.

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commongraphs - xample 1 Graph Solution This is a line in...

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