98 points! plz hurry! find the x-intercepts of the parabola with vertex (1, -9) and y intercept at (0, -6). a. (-1, 0), (3, 0) b. (-0.73, 0), (2.73, 0) c. (-1.48, 0), (2.48, 0) d. (-4.67, 0), (1.67, 0)

B. (-0.73, 0), (2.73, 0)

Step-by-step explanation:

vertex form of a parabola

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

y = a(x-1)^2 -9

substitute the point in to find a

-6 = a (0-1) ^2 -9

-6 = a *1 -9

add 9 to each side

-6+9 = a -9+9

3 =a

y = 3(x-1)^2 -9

FOIL

y = 3(x-1)(x-1) -9

y = 3(x^2-2x+1) -9

distribute

y = 3x^2-6x+3-9

combine like terms

y = 3x^2 -6x -6

factor out 3

y= 3(x^2 - 2x -2)

set = 0

0 =  3(x^2 - 2x -2)

x^2 - 2x -2 =0

using the quadratic formula

-b±sqrt(b^2 -4ac)

2a

-(-2) ± sqrt(2^2 - 4(1)(-2))

2(1)

2 ± sqrt(4+8)

2

2±sqrt(12)

2

2±2sqrt(3)

2

1±sqrt(3)

roots:  2.73,  and  -.73

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step-by-step explanation:

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i know the way

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