plane depictedParabola - GraphSquare trinomial y = a x 2 + b x + c.Known coordinatespoints A (-5; 0) and B (20; 0)- intersection pointsgiven parabola with axisOx. Point C - Intersectiongiven parabola with axisOy - located aboveaxes from. It is also known thatAngle A C B − 90 ∘ .Specify the leading coefficient of the square trinomial (i.e. the number a)

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plane depictedParabola - GraphSquare trinomial  y  =  a      x          2        +  b  x  +  c.Known coordinatespoints A (-5; 0) and B (20; 0)- intersection pointsgiven parabola with axisOx. Point C - Intersectiongiven parabola with axisOy - located aboveaxes from. It is also known thatAngle   A  C  B  −      90          ∘      .Specify the leading coefficient of the square trinomial (i.e. the number a)

Kristopher Beard · Accepted Answer

Let O be the origin of coordinates, then in a right triangle ABC aCO height Since   ∠  C  A  B  =      90          ∘        −  ∠  C  B  A  =  ∠  O  C  B, then   4  O  C  B  ∼  4  C  A  B, whence we get that OC2 = AO OB = 100 and OC = 10. On the other hand, C has coordinates (0; c). By Vieta&#039;s theorem, for a square trinomial   c  =  a  ×  1  ×  2, that is,10 = a (−5) 20, whence we obtain that a = −0, 1.Answer: -0, 1

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