Question
Asked Dec 14, 2019
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The graph of y=-4x^2-40x-106 is a parabola that opens in the direction ____. It has vertex _____, y intercept______, and ______ real roots. 

a) down, (-5,-6),-106,2

b) down,(-5,-6),-106,0

c) down, (5,-6), -106, 0

d) up, (-5,-6) -106,0

e) none of the above

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Expert Answer

Step 1

Write the parabola in standard form first. For the comple the square. 

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y=-4x²-40x-106 y=-4(x²+10x) –106 y=-4(x²+10x+25)–106+100 y=-4(x+5)²-6

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Step 2

In this parabola a=-4 is negative so it opens downwards. 

It has vertex= (h,k)= (-5,-6)

 

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У3-4(х+5)?-6 y3= у-а(x-Һ)?+k

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Step 3

For x=0, y= -106 , so y ...

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у3-4х -40х-106 У%3-4(0)°-40(0)-106 y3= y=-106

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