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=(-b)/(2a);~~y=(-\Delta)/(4a)  \Big)\\\\\\V_x=(-b)/(2a)=(-10)/(2)=\boxed{\bf-5}\\\\\\V_y=(-\Delta)/(4a)=(-40)/(4)=\boxed{\bf-10}\\\\\\\implies~\boxed{\boxed{\bf~V\Big(-5;~-10\Big)}}`