f(x)=2x+1, x greater than or equal to -1
x^2+3, x greater than -1
x^2+3, x greater than -1
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f(x) = 2x + 1
The y-intercept is 1, so (0, 1) is a point on the line.
f(-1) = -1, so (-1, -1) is another point on the line.
Draw a line starting at (-1, -1) and passing through (0, 1).
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y = x² + 3
This is an up-opening parabola with vertex (0, 3).
Draw a dotted vertical line at x = -1.
Draw the section of the parabola lying to the right of the dotted line.
The y-intercept is 1, so (0, 1) is a point on the line.
f(-1) = -1, so (-1, -1) is another point on the line.
Draw a line starting at (-1, -1) and passing through (0, 1).
:::::
y = x² + 3
This is an up-opening parabola with vertex (0, 3).
Draw a dotted vertical line at x = -1.
Draw the section of the parabola lying to the right of the dotted line.
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If these are piecewise functions then i dont think you can graph it because there are two differant funtions that describe the behavior for number greater then -1.
If these are two seperate problems then you first draw a linear graph with slope 2, y-intercept at +1, and only draw it for x values greater then -1.
The second graph is just a translation of an x^2 graph in which you move the origin from (0,0) to (0,3)
then just graph for values of x greater then -1.
If these are two seperate problems then you first draw a linear graph with slope 2, y-intercept at +1, and only draw it for x values greater then -1.
The second graph is just a translation of an x^2 graph in which you move the origin from (0,0) to (0,3)
then just graph for values of x greater then -1.
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The first is a linear function with a y-intercept of 1 and a slope of 2. So, (0,1) is one point on the line, then using a rise of two and a run of one for the slope gets points (1, 3), (2, 5), ... . Because of the restriction on the domain, you only want the portion of the graph to the right of x=-1 (that would be the point (-1, -1). Your teach would likely want a solid point put on that point because of the equal part of the domain restriction.
The second function is a quadratic. It's the standard parabola y=x^2 shifted up 3 units. The standard parabola has points (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4) etc. Shifting them up three gets (-2, 7), (-1, 4), (0, 3), (1, 4), (2, 7). Again, you only want the portion of the graph to the right of x=-1, except this time put an open circle on the end point (-1, 4) to indicate the that function cannot equal the point.
The second function is a quadratic. It's the standard parabola y=x^2 shifted up 3 units. The standard parabola has points (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4) etc. Shifting them up three gets (-2, 7), (-1, 4), (0, 3), (1, 4), (2, 7). Again, you only want the portion of the graph to the right of x=-1, except this time put an open circle on the end point (-1, 4) to indicate the that function cannot equal the point.
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You shouldn't be able to graph this, it would be a non-existing function. Are you sure you typed this right?
To D.W: This is a piecewise, to my understanding, and your steps to graphing are wrong.
To D.W: This is a piecewise, to my understanding, and your steps to graphing are wrong.