Question

Determine the zeros of the function
5) y=-x² + 2x + 1

Answer 1

zeros of the function y=
`-x^2 + 2x + 1` is
`x = 1- \sqrt {2}` &
`x = 1 + \sqrt {2}`

Explanation:

Here we have , y=
`-x^2 + 2x + 1` in order to find zeros of this quadratic function we get:
`-x^2 + 2x + 1 = 0`

⇒
`-x^2 + 2x + 1 = 0`

⇒
`x=\frac{-b \mp \sqrt{\left(b^(2)-4 a c\right.})}{2 a}`

⇒
`x = (-(2) \mp √(2^2-4(-1)(1)) )/(2(-1))`

⇒
`x = (-2 \mp √(4+4)) )/(-2)`

⇒
`x = (-2 \mp √(8) )/(-2)`

⇒
`x = 1 \mp ( √(8) )/(2)`

Since, we have two root one with positive & other with negative sign so :

⇒
`x = 1+ (2 √(2) )/(2)`

Therefore, zeros of the function y=
`-x^2 + 2x + 1` is
`x = 1- \sqrt {2}` &
`x = 1+ \sqrt {2}` .

Answer 2

Zeros of function are
`x = 1 + √(2) \text{ and } x = 1 - √(2)`

Solution:

We have to find the zeros of the function

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

Find the zeros of function:

`-x^2 + 2x+1 = 0\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=(-b\pm √(b^2-4ac))/(2a)`

`\mathrm{For\:}\quad a=-1,\:b=2,\:c=1\\\\x =(-2\pm √(2^2-4\left(-1\right)1))/(2\left(-1\right))`

`Simplify\\\\x=(-2 \pm √(4+4))/(-2)\\\\x =(-2 \pm √(8))/(-2)\\\\Simplify\\\\x =(-2 \pm 2 √(2))/(-2)\\\\x = 1 \pm √(2)`

We have two zeros

`x = 1 + √(2) \text{ and } x = 1 - √(2)`

Thus zeros of function are
`x = 1 + √(2) \text{ and } x = 1 - √(2)`