The reciprocal of a linear function and a quadratic function have the same horizontal asymptote.what is its equation?

You talk about two functions, yet ask about "its" equation.

Better clarify just what it is you want.

this is the question my teacher ask for me to do it.... help me someone

To find the equation of a reciprocal function, we need to know the equations of the original linear and quadratic functions. Let's assume the linear function is y = mx + b and the quadratic function is y = ax^2 + bx + c.

The reciprocal of a function is found by taking the reciprocal of the dependent variable (y). So, the reciprocal of y = mx + b is 1/(mx + b), and the reciprocal of y = ax^2 + bx + c is 1/(ax^2 + bx + c).

For these two reciprocals to have the same horizontal asymptote, the degrees of the numerator and denominator must be the same. Since the numerator is always 1, the denominator must also have a degree of 1.

Therefore, to find the equation, we need to set the degree of the denominator to 1 for both functions. Since the linear function already has a degree of 1, we only need to make the quadratic function have a degree of 1.

We can achieve this by factoring out an x from the quadratic equation. Suppose the quadratic function can be factored as y = a(x - r)(x - s), where r and s are the roots of the quadratic equation.

Now, the equation becomes y = a(x - r)(x - s), and the reciprocal is 1/(a(x - r)(x - s)).

To have the same horizontal asymptote, the factor a(x - r)(x - s) should not have any x-term(s) with a higher degree. Therefore, for the quadratic equation to have a degree of 1, the roots r and s must be the same, representing a single repeated root.

By setting r = s, we can rewrite the equation as y = a(x - r)^2.

Now, the reciprocal becomes 1/(a(x - r)^2), which has a horizontal asymptote at y = 0.

Therefore, the equation of the reciprocal function with the same horizontal asymptote is 1/(a(x - r)^2).