Let f(x)=x2+4x+12 . what is the vertex form of f(x)? what is the minimum value of f(x)?

well, the quadratic has a leading term with a positive coefficient, meaning is a parabola opening upwards, like a "bowl", comes from above down down down, reaches a U-turn, namely the vertex, and goes back up up up.

so the minimum value is at the vertex of course, and the minumum is well, just the y-coordinate of the vertex, 8.

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step-by-step explanation:

< 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)