We can rewrite the equation into its vertex form, as shown below
[tex]\begin{gathered} y=x^2-2x-7 \\ y=a(x-h)^2+k \\ \Rightarrow y=ax^2-2ah+ah^2+k \\ \Rightarrow\begin{cases}a=1 \\ -2ah=-2\Rightarrow h=1 \\ ah^2+k=-7\Rightarrow1+k=-7\Rightarrow k=-8\end{cases} \end{gathered}[/tex]
Therefore, the vertex form of y=x^2-2x-7 is
[tex]y=(x-1)^2-8[/tex]
On the other hand, the graph of the equation of question 1 seems to be the same graph like the one of the function y=x^2-2x-7. The y-intercept is the same (0,-7) and the vertex seems to be located on the same point (1,-8).
The two graphs are the same, only the scale changes.
These two graphs are the same.
