SOLUTION: 3x^2 + 9x, find x and y intercepts,range and vertex.

Algebra.Com
Question 375879: 3x^2 + 9x, find x and y intercepts,range and vertex.
Found 2 solutions by Fombitz, ewatrrr:
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!

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Y=intercept:

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X-intercepts:


Two solutions:

(,)
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(,)
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Convert to vertex form, by completing the square.




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Since , the parabola opens upwards and the value at the vertex is the function minimum.
Range: (,)
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Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!


Hi,

y = 3x^2 + 9x 

y intercept shen x = 0

(0,0) the y intercept

x-intercepts is when y = 0

0 = 3x^2 + 9x

0 = 3x(x+3)

x= 0

x = -3

x-intercepts (0,0) and (-3,0)

the vertex form of a parabola,  where(h,k) is the vertex

y = 3(x^2 + 3x)

y = 3[(x+ 3/2)^2 - 9/4]

y = 3(x + 3/2)^2 - 27/4

vertex Pt(-3/2,-27/4)

Range [-27/4,)

 

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