Question

Find the xintercept(s) and the coordinates of the vertex for the parabola y = -x?+2x-1. If there is more than one x-intercept, separate them with commas

Answer 1

The form of the quadratic equation is

`y=ax^2+bx+c`

The coordinates of its vertex are (h, k), where

`\begin{gathered} h=-(b)/(2a) \\ k=f(h) \end{gathered}`

The given equation is

`y=-x^2+2x-1`

Compare it with the form above

`\begin{gathered} a=-1 \\ b=2 \\ c=-1 \end{gathered}`

To find its x-intercepts, substitute y by 0

`\begin{gathered} 0=-x^2+2x-1 \\ -x^2+2x-1=0 \end{gathered}`

Multiply both sides by -1

`x^2-2x+1=0`

Factor it into 2 factors

`\begin{gathered} x^2=(x)(x) \\ 1=(-1)(-1) \\ (x)(-1)+(x)(-1)=-x-x=-2x \end{gathered}`

Then the factors are (x - 1) and (x - 1)

`\begin{gathered} x^2-2x+1=(x-1)(x-1)_{} \\ (x-1)(x-1)=0 \end{gathered}`

Equate the factor by 0 to find x

`\begin{gathered} x-1=0 \\ x-1+1=0+1 \\ x=1 \end{gathered}`

There is one x-intercept (1, 0)

To find the vertex use the rule of the vertex up

`\begin{gathered} h=-(2)/(2(-1)) \\ h=-(2)/(-2) \\ h=1 \end{gathered}`

Substitute x by 1 in the equation to find k

`\begin{gathered} k=-(1)^2+2(1)-1 \\ k=-1+2-1 \\ k=0 \end{gathered}`

The coordinates of the vertex are (1, 0)