Question

(02.04 LC)

What is the equation of the following graph in vertex form?

Courtesy of Texas Instruments
Oy=(x-1)²
Oy=(x-1)²+1
Oy=(x + 1)²-1
(0, 1)

Answer 1

first off let's take a looksie at the picture above, hmmm the vertex is at (-1 , 0) and it has another point at (0 , 1) ok hmmm

`~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{a~is~negative}{op ens~\cap}\qquad \stackrel{a~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} h=-1\\ k=0\\ \end{cases}\implies y=a(~~x-(-1)~~)^2 + 0\hspace{4em}\textit{we also know that} \begin{cases} x=0\\ y=1 \end{cases}`

`1=a(~~0-(-1)~~)^2 + 0\implies 1=a(1)^2\implies 1=a \\\\\\ y=1(~~x-(-1)~~)^2 + 0\implies {\Large \begin{array}{llll} y=(x+1)^2 \end{array}}`