Respuesta :

Answer:

[tex]\tt f(x)=2x^2+20x-10[/tex]

let y=f(x)

[tex]\tt y=2x^2+20x-10[/tex]

[tex]\tt 2x^2+10x=y+10[/tex]

[tex]\tt x^2+10x=\frac{1}{2} y+5[/tex]

[tex]\tt x^2+10x+25=\frac{1}{2} y+30[/tex]

[tex]\tt (x+5)^2=\frac{1}{2} (y+60)[/tex]

[tex]\tt 2(x+5)^2=f(x)+60[/tex]

[tex]\boxed{\tt f(x)=2(x+5)^2-60}[/tex]

Hope it helps! :)

Answer:

nope, sorry

Step-by-step explanation: