Determine the equation of a function f(X)

[math_fun]

I have to find a possible equation for which the following is true: [MATH]f(5)=0[/MATH], [MATH]f(-\sqrt{-2})=0[/MATH], there is a point of inflection at (2,0) and f(X) has a y intercept of 6.

My answer is [MATH]f(X)=(x-5)(x-2)^5(x+i\sqrt{2})[/MATH].

The given answer is [MATH]f(X)=(x-5)(x-2)^5(x^2+2)[/MATH], where the factor with complex roots has been transformed to have a degree of 2. In this case [MATH]x= +/- \sqrt{-2}[/MATH] where x is both positive and negative. In my answer [MATH]x=-\sqrt{-2}[/MATH], where x is the negative value (as stated in the question).
Is the given answer wrong, or am I wrong? And could you explain why? Thank you so much!

 

[Dr.Peterson]

Their answer has real coefficients, which is an implicit requirement given that they ask about the graph. To have real coefficients, you have to include the conjugate of the non-real zero.

Yours meets some of the requirements; but what is the y-intercept? How can you graph it to find a point of inflection?

On the other hand, their answer doesn't have f(0) = 6, either. Are you sure that's the answer for this problem? Did you leave something out?

Note also that they only ask for "a possible equation". There are many! Why did you (or they) pick 5 as an exponent, for example? The important thing is merely to check whether your answer meets the requirements.

 

[Steven G]

Complex roots comes in conjugate pairs. That is if a+bi is a root then a-bi is a root IF the function is to have real coefficients.

 

[Steven G]

Another thing, X and x are NOT the same so you can't write things like f(X) = x^2 and think that f(x) = 25. It doesn't. f(5) = x^2 as x^2 is a constant wrt to X