Respuesta :
The vertex form is [tex]y=(x-1)^{2} -6[/tex]
Step-by-step explanation:
The given equation [tex]y=x^{2} -2x-5[/tex] is a parabola. The vertex of the parabola is of the form [tex]y=ax^{2} +bx+c[/tex]
The parameters of the parabola are [tex]a=1, b=-2[/tex] and [tex]c=-5[/tex]
The vertex of the x-coordinate is given by [tex]x=\frac{-b}{2a}[/tex]
Thus, substituting the values of a and b in [tex]x=\frac{-b}{2a}[/tex], we get,
[tex]\begin{aligned}x &=\frac{-(-2)}{2(1)} \\&=\frac{2}{2} \\x &=1\end{aligned}[/tex]
Now, substituting [tex]x=1[/tex] in [tex]y=x^{2} -2x-5[/tex], we get,
[tex]\begin{aligned}y &=1^{2}-2(1)-5 \\&=1-2-5 \\y &=-6\end{aligned}[/tex]
Thus, the vertex is [tex](1,-6)[/tex]
The equation of parabola to be in vertex form is [tex]y=a(x-h)^{2} +k[/tex]
where a is the coefficient of [tex]x^{2}[/tex], h is vertex of the x-coordinate and k is the vertex of the y-coordinate.
Hence, [tex]a=1, h=1[/tex] and [tex]k=-6[/tex]
Thus, substituting the values in the equation of the vertex form,
[tex]y=1(x-1)^{2} -6\\y=(x-1)^{2} -6[/tex]
Hence, the vertex form is [tex]y=(x-1)^{2} -6[/tex]
