# plane depicted Parabola - Graph Square trinomial y=ax^(2)+bx+c. Known coordinates points A (-5; 0) and B (20; 0) - intersection points given parabola with axis Ox. Point C - Intersection given parabola with axis Oy - located above axes from. It is also known that Angle ACB-90^(circ). Specify the leading coefficient of the square trinomial (i.e. the number a)

plane depicted
Parabola - Graph
Square trinomial
$y=a{x}^{2}+bx+c$.
Known coordinates
points A (-5; 0) and B (20; 0)
- intersection points
given parabola with axis
Ox. Point C - Intersection
given parabola with axis
Oy - located above
axes from. It is also known that
Angle $ACB-{90}^{\circ }$.
Specify the leading coefficient of the square trinomial (i.e. the number a)
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Kristopher Beard
Let O be the origin of coordinates, then in a right triangle ABC a
CO height Since $\mathrm{\angle }CAB={90}^{\circ }-\mathrm{\angle }CBA=\mathrm{\angle }OCB$, then $4OCB\sim 4CAB$, whence we get that OC2 = AO OB = 100 and OC = 10. On the other hand, C has coordinates (0; c). By Vieta's theorem, for a square trinomial $c=a×1×2$, that is,
10 = a (−5) 20, whence we obtain that a = −0, 1.