Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Basic Polynomial Graphs - Problem 1

Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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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.