# How do you find the vertex and the intercepts for k(x) = -2(x² + 2x) + 1?

#### Explanation:

The given function:

$k \left(x\right) = - 2 \left({x}^{2} + 2 x\right) + 1$

$y = - 2 \left({x}^{2} + 2 x + 1\right) + 2 + 1$

$y = - 2 {\left(x + 1\right)}^{2} + 3$

${\left(x + 1\right)}^{2} = - \frac{1}{2} \left(y - 3\right)$

The above equation shows a downward parabola: ${\left(x - {x}_{1}\right)}^{2} = - 4 a \left(y - {y}_{1}\right)$ with

vertex at $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(- 1 , 3\right)$

x-intercept: setting $k \left(x\right) = 0$ in given equation to get x-intercept as follows

$0 = - 2 \left({x}^{2} + 2 x\right) + 1$

$2 {x}^{2} + 4 x - 1 = 0$

$x = \setminus \frac{- 4 \setminus \pm \setminus \sqrt{{4}^{2} - 4 \left(2\right) \left(- 1\right)}}{2 \left(2\right)}$

$x = - 1 \setminus \pm \setminus \sqrt{\frac{3}{2}}$

hence the x-intercepts are $x = - 1 \setminus \pm \setminus \sqrt{\frac{3}{2}}$

y-intercept: setting $x = 0$ in given equation to get y-intercept as follows

$k \left(x\right) = - 2 \left({0}^{2} + 2 \setminus \cdot 0\right) + 1$

$k \left(x\right) = 1$

hence the y-intercept is $y = 1$