How do I get the equation of a parabola?

There are many equation for a parabola.  Which one do they want you to solve?

good luck

zeros= roots = when x=0

o yeh.. where ever the graph cuts the x axis.

Usually you would have a equation, say:

ax^2 + bx +c  
    Or      
 (x + a) (x + b)    [plus or minus.. is doesn't really matter]

For your question it seems like you haven't been given the equation of the parabola. If that's the case, then you must be given then drawing? If it is just the drawing you can just look at where the graph cuts the x-axis. Simple? If that is not what you're trying to ask then.. you need to explain more.

well its complicated. but here's the formula for the axis of symetery: X= -b-4(a)(c)/2(b)once you figure that out you plug it into: X= -2(a)/b

/=over or the fraction bar

hope this helped.

Yes, before you can solve an equation, you must first have an equation.

The form of a quadratic equation is y = ax² + bx + c, where a, b and c are exponents and c is the y-component of the y-intercept, (x, c).

To calculate the zeros, or roots, set the equation to zero and solve for x. Being a quadratic, or second order equation (highest exponent is 2), there will be two roots.

The quadratic graphs as a parabola.

To determine whether the parabola has a maximum or minimum and  whether it opens upwards or downwards, determine the second derivative. If it is negative, the parabola opens upward and has a minimum. If it is positive, it opens downward and has a maximum.

To determine the value of the minimum or maximum, set the first derivative to zero and solve for x. That will give you the x-component of the vertex. Sub the x-value into the equation to get the y-component.

There are other parameters of a quadratic equation such as vertex, focus, directrix and latus rectum, but to explain them all here is beyond the scope of your question.