What is the conic form of the quadratic equation f(x)=2x^2+4x+12. *?

Conic sections are also known as quadratic relations because the equations which describe them are second order and not always functions. The four basic conic sections are circle, ellipse, parabola, and hyperbola.

Differentiate between quadratic relations and quadratic functions, the general equation of a quadratic function is:

y = ax^2 + bx + c.

Consider next the relation:

y^2 = x

Naïvely one might rewrite this as: y = sqrt (x). However, we lost one branch and properly it would be written: y = ±sqrt (x), then it would be a parabola opening in the x direction. But it does not pass the vertical line test and thus is only a relation and not a function.

Keeping this in mind, we can now view a quadratic relation, which is specified by the general equation (or inequality) of the form:

Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0.

Your parabola can be expressed in this form:

2x^2 + 0xy + 0y^2 + 4x - y + 12 = 0.

Drop the zeroes and you're done:

2x^2 + 4x - y + 12 = 0.

Is that what you wanted? I haven't heard the term "Conic form of a quadratic equation" before.