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

1 Answer

See answers below

Explanation:

The given function:

k(x)=-2(x^2+2x)+1

y=-2(x^2+2x+1)+2+1

y=-2(x+1)^2+3

(x+1)^2=-1/2(y-3)

The above equation shows a downward parabola: (x-x_1)^2=-4a(y-y_1) with

vertex at (x_1, y_1)\equiv(-1, 3)

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

0=-2(x^2+2x)+1

2x^2+4x-1=0

x=\frac{-4\pm\sqrt{4^2-4(2)(-1)}}{2(2)}

x=-1\pm\sqrt{3/2}

hence the x-intercepts are x=-1\pm\sqrt{3/2}

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

k(x)=-2(0^2+2\cdot 0)+1

k(x)=1

hence the y-intercept is y=1