Chapter 4 Quadratic Functions and Factoring - 4.10 Write Quadratic Functions and Models - 4.10 Exercises - Skill Practice - Page 313: 43

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The standard form of the equation is: $y=a(x-h)^2+k$, where $(h,k)$ is the vertex. Since the vertex is $(0,-1)$, we get $h=0,k=-1$. Substitute: $y=a(x-0)^2+(-1)\\ \rightarrow y=ax^2-1$ Substitute $(-2,3)$ $3=a(-2)^2-1\\ \rightarrow a=1$ Hence, the vertex form is $y=x^2-1$ The standard form: $y=ax^2+bx+c$ Since $a=1\\b=0\\c=-1$ We have: $y=x^2-1$ The intercept form of the equation: $y=a(x-p)(x-q)\\ \rightarrow y=(x-1)(x+1)$