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

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