Question

Find the vertex V and the x-intercepts x1 and x2 of the quadratic function f(x) = 2x2 - 8x.

Answer 1

`f(x)=2x^2-8x=2x(x-4)\\\\x-intercept\to f(x)=0\\\\2x(x-4)=0\iff2x=0\ \vee\ x-4=0\\\\\boxed{x_1=0\ \vee\ x_2=4}`

`\text{The vertex of}\ f(x)=ax^2+bx+c:\\\\(h,\ k)\to h=(-b)/(2a),\ k=f(h)\\\\\text{We have}\ f(x)=2x^2-8x\\\\a=2,\ b=-8,\ c=0\\\\\text{substitute}\\\\h=(-(-8))/(2\cdot2)=(8)/(4)=2\\\\k=f(2)=2(2^2)-8(2)=2(4)-16=8-16=-8\\\\\boxed{The\ vertex\ (2,\ -8)}`