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

Question

1122 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-28T15:37:57

See answers below

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#