Test 4 Section 3 #13

You’re told that the given parabolas intersect at (k,0) and (-k,0). Important insight: when two graphs intersect at a point, that point is on both graphs. Therefore, we don’t need to use both equations to solve this question, we can just pick one and find the values of x that will result in a function value of zero.

f(x)=8x^2-2

f(k)=8k^2-2

0=8k^2-2

2=8k^2

\dfrac{1}{4}=k^2

\pm\dfrac{1}{2}=k

The given figure tells you that k is positive (look where k and -k are on the x-axis) so you know the answer is \dfrac{1}{2}.

Note: You can also backsolve this one. Put the answer choices into one of the functions until you get a result of 0. Even without a calculator, that won’t take long (but will probably take longer than the algebra unless you’re stuck).

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