How do you find the vertex and the intercepts for #g(x)=x^2+2x-3 #?

1 Answer
Binayaka C.
Oct 4, 2016

Vertex at #(-1,-4)#, y-intercept is at #(0,-3)#, x-intercept is at #(-3.0) and (1,0)#

Explanation:

#g(x)=y= x^2+2x-3 = (x^2+2x+1) -4 =(x+1)^2-4# Comparing with the general equation in vertex form #a(x-h)^2-k#, we get vertex at #(-1,-4)#
To find y-intercept putting #x=0# in the equation we get #y=0+0-3=-3#. So y-intercept is at #(0,-3)#
To find x-intercept putting #y=0# in the equation we get #x^2+2x-3=0 or (x+3)(x-1)=0 :. x= -3 or x =1#. So x-intercept is at #(-3.0) and (1,0)# graph{x^2+2x-3 [-10, 10, -5, 5]}[Ans]

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
1171 views around the world
You can reuse this answer
Creative Commons License