Question

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

Answer 1

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

Step-by-step explanation:

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

⇒ $-x^2 + 2x + 1 = 0$

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

⇒ $x = \frac{-(2) \mp \sqrt{2^2-4(-1)(1)} }{2(-1)}$

⇒ $x = \frac{-2 \mp \sqrt{4+4)} }{-2}$

⇒ $x = \frac{-2 \mp \sqrt{8} }{-2}$

⇒ $x = 1 \mp \frac{ \sqrt{8} }{2}$      

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

⇒ $x = 1+ \frac{2 \sqrt{2} }{2}$

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 $x = 1 + \sqrt{2} \text{ and } x = 1 - \sqrt{2}$

Solution:

We have to find the zeros of the function

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

Find the zeros of function:

$-x^2 + 2x+1 = 0\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-1,\:b=2,\:c=1\\\\x =\frac{-2\pm \sqrt{2^2-4\left(-1\right)1}}{2\left(-1\right)}$

$Simplify\\\\x=\frac{-2 \pm \sqrt{4+4}}{-2}\\\\x =\frac{-2 \pm \sqrt{8}}{-2}\\\\Simplify\\\\x =\frac{-2 \pm 2 \sqrt{2}}{-2}\\\\x = 1 \pm \sqrt{2}$

We have two zeros

$x = 1 + \sqrt{2} \text{ and } x = 1 - \sqrt{2}$

Thus zeros of function are $x = 1 + \sqrt{2} \text{ and } x = 1 - \sqrt{2}$