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Question 449161: Is there an alternative way to graphing functions such as y=x^2+2x and x^3+x^2-2x (just by looking/observing the function), besides making the traditional x and y chart? I'm familiar with graphing the traditional parabolas and functions just by looking at the problem, but the extra degrees and x's completely threw me off and I'm not sure where to start. Thank you in advance for your help! It's greatly appreciated!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Is there an alternative way to graphing functions such as y=x^2+2x and x^3+x^2-2x (just by looking/observing the function), besides making the traditional x and y chart? I'm familiar with graphing the traditional parabolas and functions just by looking at the problem, but the extra degrees and x's completely threw me off and I'm not sure where to start. Thank you in advance for your help! It's greatly appreciated!
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Yes, there much more efficient ways to graph equations of first and second degree. First degree equations are straight lines that have the following standard form: y=mx+b, with m=slope, and b=y-intercept. You can always write the equation of a straight line given the slope and a point (x,y) coordinates on the line or given two points because you can find the slope by calculating ∆y/∆x. To find b, the y-intercept, just plug in the (x,y) coordinates and slope in the standard form and solve for b. To graph the equation, all you need to do is connect the y-intercept to a given point on the line and as a check, the slope of the line must match what you got initially. Another check is the x-intercept which you can find by setting y=0, then solving for x, given the slope and y-intercept. This seems like more work than it is, but it isn't. The best thing you can do is become familiar with the standard form for a straight line:y=mx+b
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All second degree equations are parabolas in one form or another. Again, becoming familiar with the standard form is the best way to recognize what the graph would look like without plotting x's and y's. One familiar standard form of a parabola is: y=(x-h)^2+k, where (h,k) are the (x,y) coordinates of the vertex. For example, take the equation you submitted: y=x^2+2x. To put this into standard form you must complete the square, like so:
y=x^2+2x
y=(x^2+2x+1)-1
y=(x+1)^2-1
From this equation, you know that this is a parabola with vertex is at (-1,-1), and it opens upwards because the coefficient of the x^2 term is positive. You can see now you have a good idea what the parabola looks like on the graph. With another point like the x-intercept and knowing the axis of symmetry, you should be able to draw the complete graph. Another way of finding points to graph, is calculating x and y intercepts. For example, the second equation you submittted, x^3+x^2-2x, you would be calculating the x-intercepts which are points on the graph. Of course, there is much more to know about shifting and bumping curves, but there is a better way to graph equations than plotting x's and y's.
Hope this brief explanation helps.
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