Use the quadratic formula to determine the x-intercepts (if any)f(x)=2x^2=2
am I right so far?
f(x)ax^2+bx+c
y=ax^2+bx+c
y=2x^2+2
y=0=>2x^2+2=0
=>2x^2+2=0
ax^2=2,bx=2,c=0
x=b=sqrtb^2+4ac over 2a
x=(2)+sqrt(2)^24(2)(0)over2(2)
x=sqrt4+4over 2
No, No.
2x^2+ 0x + 2
a=2, b=0, c=2