Question 213488
You're on the right track. A function is one-to-one if {{{f(a)=f(b)}}} <u>only</u> means {{{a=b}}}. In other words, if two outputs of a function are equal, then the inputs <u>must</u> be equal for the function to be one-to-one.



So if you think that this function is NOT one-to-one, then simply pick a counter-example to prove this claim false. Let's say that {{{f(a)=f(b)=-3}}}. Plugging this into {{{f(x)=x^2+2x-3}}} gets us {{{-3=x^2+2x-3}}}. Now solve for 'x' to get {{{x=0}}} or {{{x=-2}}}



So if we plug in either 0 or -2 into f(x), we'll get -3 as an output. Ie. {{{f(0)=f(-2)=-3}}}. Since {{{0<>-2}}}, this means that the function is NOT one-to-one. So you are correct.