Question 165212
Let's evaluate the left endpoint a=2



{{{f(x)=3x^2-2x-11}}} Start with the given equation.



{{{f(2)=3(2)^2-2(2)-11}}} Plug in {{{x=2}}}.



{{{f(2)=3(4)-2(2)-11}}} Square {{{2}}} to get {{{4}}}.



{{{f(2)=12-2(2)-11}}} Multiply {{{3}}} and {{{4}}} to get {{{12}}}.



{{{f(2)=12-4-11}}} Multiply {{{-2}}} and {{{2}}} to get {{{-4}}}.



{{{f(2)=-3}}} Combine like terms.



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Let's evaluate the right endpoint b=3



{{{f(x)=3x^2-2x-11}}} Start with the given equation.



{{{f(3)=3(3)^2-2(3)-11}}} Plug in {{{x=3}}}.



{{{f(3)=3(9)-2(3)-11}}} Square {{{3}}} to get {{{9}}}.



{{{f(3)=27-2(3)-11}}} Multiply {{{3}}} and {{{9}}} to get {{{27}}}.



{{{f(3)=27-6-11}}} Multiply {{{-2}}} and {{{3}}} to get {{{-6}}}.



{{{f(3)=10}}} Combine like terms.



So as x changes from 2 to 3, f(x) (ie y) changes from -3 to 10 which means that the graph MUST have crossed over the x-axis somewhere in between x=2 and x=3. So this shows that there is a zero between a and b