Edward C. answered • 01/26/16
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Caltech Grad for math tutoring: Algebra through Calculus
When you don't know how to graph a function the easiest thing to do is to try plotting points. This means that you pick a value for x, figure out what y is equal to for that value of x, and then plot the point (x,y). Keep doing this until you get a shape that you recognize or you can figure out how the graph must continue for other values of x (both positive and negative). To make things easy try picking points for x that make y easy to evaluate. In this case small integer values of x are easy to calculate. For example,
x = 0 ==> y = 1 - 0/6 = 1 ==> the point (0,1) is on the graph
x = 1 ==> y = 1 - 1/6 = 5/6 ==> the point (1,5/6) is on the graph
x = -1 ==> y = 1 - 1/6 = 5/6 ==> the point(-1,5/6) is on the graph
Try plotting a few more points (like x = ±2, x = ±3) and see if you get a shape that you recognize. I don't know what level of math you are in so I don't know if you have learned certain things yet like how to find the vertex of a parabola, whether the parabola opens up or down, how to find the axis of symmetry, and so on, so I don't want to confuse you with too much information. Feel free to comment back if you have questions.
Edward C.
tutor
Well you still need to calculate a few values so you know where to plot the points on the graph. But if you use some of the properties of quadratics you can get by with just evaluating 3 or 4 points instead of 5 or 10.
First write the quadratic in standard form y = ax2 + bx + c. In this case that is y = -(1/6)x2 + 1, so a = -(1/6), b = 0 and c = 1. The axis of symmetry (AOS) is the vertical line x = -b/(2a) which in this case is x = -0/(-2/6) so the AOS is the line x = 0 which is just the y-axis. The graph will be symmetrical about this line so you only need to find points on one side of the line and then reflect them over the line to the other side. For example, as we found above the point (1,5/6) is on the graph so the symmetrical point (-1,5/6) is also on the graph.
The vertex lies on the AOS and is found by plugging in the x value from the AOS (which in this case is 0) into the equation, so it is the point (0,1) that we found above. Since a = -(1/6) is negative the parabola will open downward, which means that the vertex is the maximum value of the graph.
Other useful points to find are the x-intercepts, which are the values of x that will make y = 0. So set y = 0 and solve the equation -(1/6)x2 + 1 = 0 ==> -(1/6)x2 = -1 ==> x2 = 6 ==> x = ±√6 ≅ ±2.4. So the points (2.4,0) and (-2.4,0) are also on the graph. Again, note the symmetry about the AOS.
The next x-value that will make y an integer is x = 6 which gives y = 1 - 62/6 = 1 - 6 = -5. So the point (6,-5) is on the graph, and by symmetry the point (-6,-5) is also on the graph.
If you plot these 7 points on graph paper (and remember, we only had to actually calculate 4 of them) you should be able to draw a smooth curve thru them of a broad upside down bowl shape. Draw arrows pointing down and out on each side of the graph to indicate that it continues on in each direction. By the way, the curve is broad (or spread out wide) because the absolute value of the "a" coefficient is less than 1.
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01/26/16
Pia R.
This a big help!! Thank you so much, Edward! :-)
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01/27/16
Pia R.
01/26/16