How do you find the vertex and the intercepts for g(x)=x^2+2x-3 g(x)=x2+2x−3?

1 Answer
Binayaka C.
Oct 4, 2016

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

Explanation:

g(x)=y= x^2+2x-3 = (x^2+2x+1) -4 =(x+1)^2-4g(x)=y=x2+2x−3=(x2+2x+1)−4=(x+1)2−4 Comparing with the general equation in vertex form a(x-h)^2-ka(x−h)2−k, we get vertex at (-1,-4)(−1,−4)
To find y-intercept putting x=0x=0 in the equation we get y=0+0-3=-3y=0+0−3=−3. So y-intercept is at (0,-3)(0,−3)
To find x-intercept putting y=0y=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]

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