How do you find the x-intercept of y=x^2+2x-3? I know you set it to 0 but I am not sure what to do after that

Question

zepp · Answer

You either factor it out or use the quadratic formula ;)

anonymous · Answer

oh ok thank you :)

lgbasallote · Answer

"set it to zero" you do mean set y to zero right?

anonymous · Answer

you can factor it out

anonymous · Answer

yes thats what i meant

anonymous · Answer

you quadratic formula when u can factor out the numebers

anonymous · Answer

oh ok thank you

anonymous · Answer

cant*

anonymous · Answer

so if y = -3, and x = 1, how do you find the vertex?  or whats the formula to find it

zepp · Answer

To find the vertex, use a point and zero
Then plug in it
$\large y=a(x-z_{1})(x-z_{2})$

zepp · Answer

Uh, forget what I said. :s

anonymous · Answer

@zepp  lol ok i looked at it and was like uh...

zepp · Answer

But what's the (1,-3)?

zepp · Answer

A random point on the parabola?

callisto · Answer

Make it into the form: y=a(x-h)+k, where (h,k) are the coordinates of the vertex

anonymous · Answer

that was the first part of the problem was to find the y intercept which turned to be -3, and x intercept was 1 @zepp

zepp · Answer

A quadratic function must have 2 x-intercept or none.

zepp · Answer

There's 1 missing :)

anonymous · Answer

than the other is -3

anonymous · Answer

those were the two that i got

callisto · Answer

I suppose it's called completing of square...
y=x^2+2x-3
  = (x^2 + 2x +1) -1 -3
  = (x+1)^2 - 4
Coordinates of vertex are (-1, -4)

zepp · Answer

..Alright, the equation was given. Thought it wasn't >.>