Graph of a parabola

Question

The equation of the following graph is $-x^2-4x-c=y$ how to find c if 3OB=AO

Answer 1

Finding the roots you get $$-2\pm\sqrt{4-c}$$ then you know $$x_A=-2-\sqrt{4-c}$$$$x_B=-2+\sqrt{4-c}$$ then you need to have $x_B>0 \implies c<0$ so you need to solve $|x_A|=3x_B$ $$2+\sqrt{4-c}=3(-2+\sqrt{4-c})$$ that give you $$c=-12$$

Answer 2

Let $A(x_2,0)$ and $B(x_1,0).$

Thus, $$x_1x_2=c,$$ $$x_1+x_2=-4$$ and $$3x_1=-x_2,$$ which gives $$x_2=-6,$$ $$x_1=2$$ and $$c=-12.$$

Answer 3

Slight variation of previous answer: $$y=-x^2-4x-c=-(x-A)(x-B)=-(x-(-3B))(x-B)=-x^2-2Bx+3B^2$$ Equating the coefficients: $$-4=-2B \Rightarrow B=2; c=3B^2=12.$$