Respuesta :
an x-intercept is namely a "solution" or "zero" or "root" often called, and when that happens, y = 0, just like with any other x-intercept.
[tex]\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \boxed{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\ -------------------------------\\\\ \begin{cases} h=3\\ k=-2 \end{cases}\implies y=a(x-3)^2-2 \\\\\\ \textit{we also know that } \begin{cases} x=0\\ y=7 \end{cases}\implies 7=a(0-3)^2-2 \\\\\\ 9=9a\implies \cfrac{9}{9}=a\implies 1=a\qquad therefore\qquad \boxed{y=(x-3)^2-2}[/tex]
so what is its x-intercept anyway?
[tex]\bf \stackrel{y}{0}=(x-3)^2-2\implies 2=(x-3)^2\implies \pm\sqrt{2}=x-3 \\\\\\ \pm\sqrt{2}+3=x\qquad therefore\qquad (\pm\sqrt{2}+3~~~~,~~~~0)[/tex]
[tex]\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \boxed{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\ -------------------------------\\\\ \begin{cases} h=3\\ k=-2 \end{cases}\implies y=a(x-3)^2-2 \\\\\\ \textit{we also know that } \begin{cases} x=0\\ y=7 \end{cases}\implies 7=a(0-3)^2-2 \\\\\\ 9=9a\implies \cfrac{9}{9}=a\implies 1=a\qquad therefore\qquad \boxed{y=(x-3)^2-2}[/tex]
so what is its x-intercept anyway?
[tex]\bf \stackrel{y}{0}=(x-3)^2-2\implies 2=(x-3)^2\implies \pm\sqrt{2}=x-3 \\\\\\ \pm\sqrt{2}+3=x\qquad therefore\qquad (\pm\sqrt{2}+3~~~~,~~~~0)[/tex]
