How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry of #y=(x+2)(x-1)#?

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Answer 1 · 2017-05-01T16:45:01

The standard form for the equation of a parabola of this type is:

#y = ax^2 + bx + c" [1]"#

If #a > 0# then the parabola opens up.
If #a < 0# then the parabola opens down.

#y=(x+2)(x-1)" [2]"#

I shall use the F.O.I.L method to multiply the factors of equation [2] and then write the results in the standard form of equation [1]:

#y = x^2 -x + 2x - 2#

#y = x^2 + x - 2" [3]"#

Equation [3] is in standard form and we observe that #a = 1, b=1 and c = -2#.

#a > 0#, therefore, the parabola opens up

For this type of parabola, the equation of the axis of symmetry is:

#x = -b/(2a)#

Substituting in the values:

#x = -1/(2(1))#

#x = -1/2# is the equation of the axis of symmetry.

The same equation is used to find, h, the x coordinate of the vertex:

#h = -1/2#

The y coordinate of the vertex, k, is equal to the function evaluated at h:

#k = h^2+ h- 2#

#k = (-1/2)^2 + (-1/2)-2#

#k = -9/4#

The vertex is #(-1/2,-9/4)#