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

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: 3x^2 + 9x, find x and y intercepts,range and vertex.      Log On


   

Question 375879: 3x^2 + 9x, find x and y intercepts,range and vertex.
Found 2 solutions by Fombitz, ewatrrr:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
y=3x%5E2%2B9x
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.
Y=intercept:
x=0%7D%7D%0D%0A%7B%7B%7By=3%280%29%2B9%280%29=0
(0,0)
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.
X-intercepts:
3x%5E2%2B9x=0
3x%28x%2B3%29=0
Two solutions:
x=0
(0,0)
.
.
.
x%2B3=0
x=-3
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.
(-3,0)
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.
.
Convert to vertex form, y=a%28x-h%29%5E2%2Bk by completing the square.
y=3x%5E2%2B9x
y=3%28x%5E2%2B3x%29
y=3%28x%5E2%2B3x%2B9%2F4%29-3%289%2F4%29
y=3%28x%2B3%2F2%29%5E2-27%2F4
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Since a=3%3E0, the parabola opens upwards and the value at the vertex is the function minimum.
Range: (-27%2F4,infinity)
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Answer by ewatrrr(24785) About Me  (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, y=a%28x-h%29%5E2+%2Bk 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,infinity)