SOLUTION: rewrite function f(x)=-1/3x2-2x-9 in the form f(x)=a(x-h)2+k give x and y intercepts give domain and range

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Question 694630: rewrite function f(x)=-1/3x2-2x-9 in the form f(x)=a(x-h)2+k
give x and y intercepts
give domain and range
Found 2 solutions by MathLover1, stanbon:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=-%281%2F3%29x%5E2-2x-9 in the form f%28x%29=a%28x-h%29%5E2%2Bk
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=-0.333333333333333+x%5E2-2+x-9 Start with the given equation



+x-intercept ......set f%28x%29=0
0=-%281%2F3%29x%5E2-2x-9...use quadratic formula to solve for x

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+



x+=+%282+%2B-+sqrt%284-12+%29%29%2F%28-2%2F3%29%29+

x+=+%282+%2B-+sqrt%28-8+%29%29%2F%28-2%2F3%29%29+

x+=+%282+%2B-+2.83%2Ai%29%2F%28-2%2F3%29%29+

x+=+3%282+%2B-+2.83%2Ai%29%2F-2+

x+=+%286+%2B-+8.49%2Ai%29%2F-2+
x+=+%286%2F-2+%2B-+8.49%2Ai%2F-2%29+

x+=+-3+%2B-+4.245%2Ai+....so, we have two complex solutions, it means there is no real solutions and there is no +x-intercepts

and y-intercept......set x=0
f%28x%29=-%281%2F3%290%5E2-2%2A0-9
f%28x%29=0-0-9
f%28x%29=-9.......y-intercept is at (0,-9)

see it on a graph:
+graph%28+600%2C+600%2C+-10%2C10%2C+-15%2C+10%2C+-%281%2F3%29x%5E2-2x-9%29+

domain:



range:




Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
rewrite function
f(x)=-1/3x2-2x-9 in the form f(x)=a(x-h)2+k
y = (-1/3)(x^2 - [(-2)/(-1/3)]x + ?) - 9 -(-1/3)(?)
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Complete the square:
y = (-1/3)(x^2 +(6)x + 9) - 9 -(-1/3)9
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Factor:
y = (-1/3)(x+3))^2 - 9 + 3
----
y = (-1/3)(x+3))^2 - 6
------------
give x and y intercepts
x-intercept:: ?
Let y = 0 ; Use the Quadratic Formula to solve for "x".
-----
y-intercept:: ?
Let x = 0, then y = -9
-----
Domain: All Real Numbers
---
Range = ?
Vertex is at ((-3),-6)
Range is All Real Numbers <= -6
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Cheers,
Stan H.
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