Respuesta :
[tex]\bf ~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{ope ns~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\bf \stackrel{\textit{vertex}}{(3,-2)}~~ \begin{cases} h = 3\\ k = -2 \end{cases}\implies y=a(x-3)^2-2 \\\\\\ \stackrel{\textit{passes through}}{(5,-1)}~~ \begin{cases} x=5\\ y= -1 \end{cases}\implies -1=a(5-3)^2-2\implies 1=a(2)^2 \\\\\\ 1=4a\implies \cfrac{1}{4}=a~\hfill \boxed{y=\cfrac{1}{4}(x-3)^2-2}[/tex]
The equation of the parabola is [tex]y=\dfrac{1}{4}(x-3)^2-2[/tex].
Given:
The vertex of the parabola is [tex](3,-2)[/tex].
The parabola passes through the point [tex](5,-1)[/tex].
To find:
The equation of the parabola in vertex form.
Explanation:
The vertex form of a parabola is:
[tex]y=a(x-h)^2+k[/tex]
Where, [tex]a[/tex] is a constant and [tex](h,k)[/tex] is vertex.
The vertex of the parabola is [tex](3,-2)[/tex]. So, [tex]h=3,k=-2[/tex].
[tex]y=a(x-3)^2+(-2)[/tex]
[tex]y=a(x-3)^2-2[/tex] ...(i)
The parabola passes through the point [tex](5,-1)[/tex]. Putting [tex]x=5,y=-1[/tex] in (i), we get
[tex]-1=a(5-3)^2-2[/tex]
[tex]-1+2=a(2)^2[/tex]
[tex]1=4a[/tex]
[tex]\dfrac{1}{4}=a[/tex]
Substituting [tex]a=\dfrac{1}{4}[/tex] in (i), we get
[tex]y=\dfrac{1}{4}(x-3)^2-2[/tex]
Therefore, the equation of the parabola is [tex]y=\dfrac{1}{4}(x-3)^2-2[/tex].
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