Question

Anyone have any idea? i dont have any idea

Answer 1

Answer:

$\tt f(x)=2x^2+20x-10$

let y=f(x)

$\tt y=2x^2+20x-10$

$\tt 2x^2+10x=y+10$

$\tt x^2+10x=\frac{1}{2} y+5$

$\tt x^2+10x+25=\frac{1}{2} y+30$

$\tt (x+5)^2=\frac{1}{2} (y+60)$

$\tt 2(x+5)^2=f(x)+60$

$\boxed{\tt f(x)=2(x+5)^2-60}$

Hope it helps! :)