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# halp? pls?

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ok:

state the axis of symmetry of the parabola: y=3(x-2)^2-4

also state the vertex, and whether it is a maximum or a minimum, and the y intercept of the parabola and the graph

May 30, 2019
edited by ProffesorNobody  May 30, 2019

#1
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y = 3(x - 2)^2 - 4

Vertex  = (2, - 4)

Axis of symmetry :  x  = 2

This parabola opens upward....therefore...the vertex is a minimum

y intercept : let x  = 0   and we have

y = 3(0-2)^2 - 4   =  3(-2)^2 - 4  =  3(4) - 4  = 12 - 4  = 10

Here's the graph :   https://www.desmos.com/calculator/w1tqaeu5yv   May 31, 2019
#2
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can you explain?

ProffesorNobody  May 31, 2019
#3
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can you explain the process?

ProffesorNobody  May 31, 2019
#4
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y = 3( - 2)^2  + ( - 4 )

We have the form

y = a(x - h)^2 + k

When "a"  is positive, the parabola opens upward

The vertex  = (h, k)  = (2, -4 )

The axis of symmery  is  x  = h

The y intercept is found just as I described

The vertex is always the minimum when a parabola opens upward   CPhill  May 31, 2019
#5
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godddd thank youuuuu

ProffesorNobody  May 31, 2019