Determine the critical points and draw the graph of y=2x^2 ?

1 Answer
Apr 5, 2017

Full explanation given

Explanation:

color(blue)("Things that are useful to know")

To force maths formatting you use the hash symbol at the beginning and end of the maths bit.

So hash y=2x^2 hash gives y=2x^2
Have a look at https://socratic.org/help/symbols
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Write this as: y=2x^2+0x+0

This is a sort of cheat way of deciding how the graph looks:
Compare to the standard form of: y=ax^2+bx+c

If a is positive the graph is of shape uu
If a is negative the graph is of shape nn

The b from bx controls where the axis of symmetry is (the centre line). The axis of symmetry is at: (-1/2)xxb/a

Incidentally; the x value of the vertex (turning point) is the same as that for the axis of symmetry.

The y-intercept (where the graph crosses the y-axis) is c
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color(blue)("Answering the question: Critical points")

color(brown)("The "x^2" term is positive so the shape is "uu)

color(brown)("The axis of symmetry is at "(-1/2)xx0/1=0)

which is the y-axis. So the 'centre line' of the graph is the y-axis.

color(brown)("The value of "x_("vertex")->" axis of symmetry"->x=0)

To determine y_("vertex") set x as 0. So we have by substitution:
color(brown)("The value of "y_("vertex")->y=2x^2" "->" "y=2(0)^2=0)

color(brown)("Vertex "->(x,y)=(0,0)

color(brown)("The graph does not cross the x-axis")

Build a table of values. Mark the points on the paper and draw the best curve you can through those points.

Tony B

Tony B