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Basic Polynomial Graphs  Problem 1
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Graphing a polynomial function that is a quadratic; so for this particular problem we are just looking at the base quadratic function y equals x². Okay we have a chart with some numbers we are going to fill up the chart and then plot the points to see what this graph looks like.
If x is 2, we plug in 2, –2² we have 2 2 so 2 times 2 negatives cancel leaving us with 4. We plug in 1, 1² again the negatives cancel leaving us with 1, 0² is 0, 1² 1 and 2² is 4. Now we have a table and we can go and plot these corresponding points. Not going to make my graph exact because I'm not concerned with the exact position I just want to sort of get a rough idea of what this looks like.
Okay, plot in points x is 2 y is 4 over 2 we go up to 4, (1,1), (0,0), (1,1) and (2,4) connecting our dots we end up with a basic parabola. So y is equals f(x) is equals x² quadratic function will give us base graph of a parabola or just a 'U' basically looking graph.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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Sample Problems (13)
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Basic Polynomial Graphs
Problem 1 4,310 viewsGraph:
f(x) = x² 
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Problem 2 4,029 viewsGraph:
f(x) = x³ 
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Problem 3 3,815 viewsGraph:
f(x) = x⁴ 
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Problem 4 916 views 
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Problem 5 876 views 
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Problem 6 803 views 
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Problem 7 605 views 
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Problem 8 688 views 
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Problem 9 647 views 
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Problem 10 581 views 
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Problem 11 575 views 
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Problem 12 624 views 
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Problem 13 626 views
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