1) The point (2,-1) is on the graph of which parabola?
a) y=x^2 +2x+1
b) y = x² – 2x-1
c) y=x² + 2x-1
d) y=x^2- 2x +1

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

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step-by-step explanation:

given : the shape of his satellite can be modeled by   where x and y are modeled in inches.

now,the given equation   is a equation of parabola.

here, coefficient of x is positive, hence the parabola opens rightwards.

on comparing this equation with standard equation , we get

in standard equation, coordinates of focus=(0,m)

thus for given equation coordinates of focus=(0,1.5)

step-by-step explanation: