Elliptic and Parabolic Curves
Let f ( x ) be a quadratic function.
When
y 2 = − x 3 + x 2 − x + 1 and x 2 = − y 3 + y 2 − y + 1
are simultaneously graphed, there are three intersections. f ( x ) passes through all three of them, and there is one intersection where x = y = a .
Find f ( x ) in terms of x and a .
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3 solutions
The first equation is easy to graph, as it is simply y = + / − 1 − x + x 2 − x 3 The second equation is clearly just the inverse of the first, so just "flip" the first graph across the line y=x, and see that the three intersection points are (1,0), (0,1), and (a,a), where a is between 0 and 1 (approximately 0.682...)
Because of this, it is easily to eliminate three of the possible answers, even without knowing the exact value of a, simply because those functions would have the wrong shape. The only one leftover that could possibly intersect these three points is f ( x ) = − ( 1 + a ) x 2 + a x + 1
THIS IS NOT A COMPLETE SOLUTION:
Realize that the intersections with be ( 1 , 0 ) , ( 0 , 1 ) , and ( a , a ) . Then it is easy to solve for a quadratic, which is − ( 1 + a ) x 2 + a x + 1 .
You might need to change the options; see my solution above.
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Yes, my options aren't very well chosen I suppose...
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In fact, after realizing that the root of x 3 + x − 1 is greater than 2 1 , we know that the parabola must be concave so only the choice where the leading coefficient is negative is right :P
In the first two equations, replace x and y with a and you get
− a 3 − a + 1 = 0
f ( x ) must satisfy the above since it intersects the point ( a , a ) . Checking all the options, only y = − ( 1 + a ) x 2 + a x + 1 holds at x = y = a :
a = − ( 1 + a ) a 2 + a 2 + 1 ⟷ − a 3 − a + 1 = 0