Write in intercept form y=-x^2-x+6 a. y= -(x+2)(x-6) b. y=-(x-2)(x+3) c. y=-(x+2)(x+6) d. y=(-x+3)(x-4)

-x² - x + 6

First thing I would do is distribute out that negative. Your new problem is:

- (x² + x - 6)

Now, factor the inside. We need to find products of - 6 that when added together have a sum of positive 1. They are 3 and - 2.

- (x² + 3x - 2x - 6)

Group together and find your factors. I'm going to short cut and go right to them for the sake of space and time.

- (x + 3)(x - 2)

Your answer is Option B

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer