How do you find the vertex and intercepts for this equation g(x)=x^2+x-6?

Answer 1

x² + x - 6

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

x intercepts where y = 0; x = -3, x = 2
(-3, 0) (2,0)
y intercepts where x = 0; y = 6
(0, 6)

vertex: x = -b/2a = -1/2
x = -0.5 y = -6.5

Answer 2

Looks like Scott got this right, but I'd do it slightly different because I don't like memorizing formulas. Here's how I'd do this problem.

1. g(0) = -6

That's the y-intercept (0,-6).

2. Factor it: g(x) = (x+3)(x-2)

So if x = -3 or x = +2, g(x) will be zero. The x-intercepts are at (-3,0) and (2,0).

3. (And here's where I do it differently than Scott...) The vertex occurs halfway between the roots ... in other words, right at the midpoint between -3 and +2. That midpoint is at (-3+2)/2 = -1/2.

g(-1/2) = 1/4 - 1/2 - 6 = (1 - 2 - 24)/4 = -25/4 = -6 1/4

The vertex is at (-0.5, -6.25)

I hope that's the same answer Scott got.

[Edit: Looks like he just made a slight mistake calculating g(-1/2) when finding the vertex.]

Answer 3

g(x) = x^2 + x - 6
g(x) = (x + 3)(x - 2)

x = (-b)/(2a)
x = (-1)/(2(1))
x = (-1/2)

g(-1/2) = (-1/2)^2 + (-1/2) - 6
g(-1/2) = (1/4) - (1/2) - 6
g(-1/2) = (1/4) - (2/4) - (24/4)
g(-1/2) = (1 - 2 - 24)/4
g(-1/2) = (-25/4)

x-intercepts are -3 and 2
y-intercept is -6
Vertex is ((-1/2),(-25/4))

Answer 4

for this form of equation (y= ax^2+bx+c) there is aggregate answer: x=(-b(+or-) (b^2-4ac)^(1/2))/2a so the answer will be:
x=2 and 3