Can someone who has a few minutes could you please correct this for me.thank you.
Directions: Find the x-intercepts of
y= x^2+x-6
My answer: there are two x-intercepts
( (1-sqrt25)/(2),0) and ((1+sqrt25)/(2),0
x^2+x-6= (x+3)(x-2)
how did you get those roots?
i am trying to follow the book and that's one similar example of it....
isn't the answer 6?
i'm sorry, -6
*w*w*w*.*a*s*k*.*c*o*m* hey i cant help but heres a hint just go 2 (go on the words w/ the stares around them )and ask the question
y= x^2+x-6
Set y equal to zero, because when y equals zero, is where the x-intercept will be.
So you'll have:
0 = x^2 + x-6
Factor that to get:
0 = (x+3)(x-2)
This means that x=-3 and x=2. So your x-intercepts will be at those two points...(-3, 0) and (2, 0)
To find the x-intercepts of the equation y = x^2 + x - 6, follow these steps:
1. Set y equal to zero because the x-intercept is where y equals zero.
0 = x^2 + x - 6
2. Factor the quadratic equation to find the roots (values of x where the equation equals zero).
0 = (x + 3)(x - 2)
3. Set each factor equal to zero and solve for x.
x + 3 = 0 --> x = -3
x - 2 = 0 --> x = 2
The x-intercepts are the points (-3, 0) and (2, 0), where the graph intersects the x-axis.