Question

X^2+10x-y+15=0 what is the vertex of this parabola?

Answer 1

 

$\displaystyle\bf\\x^2+10x-y+15=0\\\\y=x^2+10x+15\\\\\Delta=b^2-4ac=10^2-4\cdot15=100-60=\boxed{\bf40}\\\\Vertex~is~~V\Big(x=\frac{-b}{2a};~~y=\frac{-\Delta}{4a} \Big)\\\\\\V_x=\frac{-b}{2a}=\frac{-10}{2}=\boxed{\bf-5}\\\\\\V_y=\frac{-\Delta}{4a}=\frac{-40}{4}=\boxed{\bf-10}\\\\\\\implies~\boxed{\boxed{\bf~V\Big(-5;~-10\Big)}}$