Respuesta :
The form of the quadratic equation is
[tex]y=ax^2+bx+c[/tex]The coordinates of its vertex are (h, k), where
[tex]\begin{gathered} h=-\frac{b}{2a} \\ k=f(h) \end{gathered}[/tex]The given equation is
[tex]y=-x^2+2x-1[/tex]Compare it with the form above
[tex]\begin{gathered} a=-1 \\ b=2 \\ c=-1 \end{gathered}[/tex]To find its x-intercepts, substitute y by 0
[tex]\begin{gathered} 0=-x^2+2x-1 \\ -x^2+2x-1=0 \end{gathered}[/tex]Multiply both sides by -1
[tex]x^2-2x+1=0[/tex]Factor it into 2 factors
[tex]\begin{gathered} x^2=(x)(x) \\ 1=(-1)(-1) \\ (x)(-1)+(x)(-1)=-x-x=-2x \end{gathered}[/tex]Then the factors are (x - 1) and (x - 1)
[tex]\begin{gathered} x^2-2x+1=(x-1)(x-1)_{} \\ (x-1)(x-1)=0 \end{gathered}[/tex]Equate the factor by 0 to find x
[tex]\begin{gathered} x-1=0 \\ x-1+1=0+1 \\ x=1 \end{gathered}[/tex]There is one x-intercept (1, 0)
To find the vertex use the rule of the vertex up
[tex]\begin{gathered} h=-\frac{2}{2(-1)} \\ h=-\frac{2}{-2} \\ h=1 \end{gathered}[/tex]Substitute x by 1 in the equation to find k
[tex]\begin{gathered} k=-(1)^2+2(1)-1 \\ k=-1+2-1 \\ k=0 \end{gathered}[/tex]The coordinates of the vertex are (1, 0)
