An x2 parabola will always have one vertex, but depending on the discriminant of the function (b2-4ac) the parabola will either have 2 roots (it crosses the x-axis twice), 1 repeating root (the parabola meets the x-axis at a single point), or no real roots (the parabola doesn't meet the x-axis at all)
The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?
In analytical geometry, the roots of a parabola are the x-values (if any) for which y = 0.
In a quadratic y = ax² + bx + c, the roots are where y = 0, and the parabola crosses the x-axis. The average of these two roots is the x coordinate of the vertex of the parabola.
Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a ≠0. The roots of the parabola are given by x = [-b ± sqrt(D)]/2a where D is the discriminant. Rather than solve explicitly for the coordinates of the vertex, note that the vertical line through the vertex is an axis of symmetry for the parabola. The two roots are symmetrical about x = -b/2a so, whatever the value of D and whether or not the parabola has real roots, the x coordinate of the vertex is -b/2a. It is simplest to substitute this value for x in the equation of the parabola to find the y-coordinate of the vertex, which is c - b2/2a.
A parabola is a graph of a 2nd degree polynomial function. Two graph a parabola, you must factor the polynomial equation and solve for the roots and the vertex. If factoring doesn't work, use the quadratic equation.
An x2 parabola will always have one vertex, but depending on the discriminant of the function (b2-4ac) the parabola will either have 2 roots (it crosses the x-axis twice), 1 repeating root (the parabola meets the x-axis at a single point), or no real roots (the parabola doesn't meet the x-axis at all)
The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?The answer will depend onwhat you mean by "solving a parabola". A parabola has a directrix and a focus, a turning point, 0 1 or 2 roots and so on. Which of these is "solving"?
In analytical geometry, the roots of a parabola are the x-values (if any) for which y = 0.
In a quadratic y = ax² + bx + c, the roots are where y = 0, and the parabola crosses the x-axis. The average of these two roots is the x coordinate of the vertex of the parabola.
Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a ≠0. The roots of the parabola are given by x = [-b ± sqrt(D)]/2a where D is the discriminant. Rather than solve explicitly for the coordinates of the vertex, note that the vertical line through the vertex is an axis of symmetry for the parabola. The two roots are symmetrical about x = -b/2a so, whatever the value of D and whether or not the parabola has real roots, the x coordinate of the vertex is -b/2a. It is simplest to substitute this value for x in the equation of the parabola to find the y-coordinate of the vertex, which is c - b2/2a.
I think you are talking about the x-intercepts. You can find the zeros of the equation of the parabola y=ax2 +bx+c by setting y equal to 0 and finding the corresponding x values. These will be the "roots" of the parabola.
There's the vertex (turning point), axis of symmetry, the roots, the maximum or minimum, and of course the parabola which is the curve.
-- The roots of a quadratic equation are the values of 'x' that make y=0 . -- When you graph a quadratic equation, the graph is a parabola. -- The points on the parabola where y=0 are the points where it crosses the x-axis. -- If it doesn't cross the x-axis, then the roots are complex or pure imaginary, and you can't see them on a graph.
The "solution" depends on what the question is.If you mean the roots, and if the equation of the parabola isy = ax^2 + bx + c then the roots are[-b +/- sqrt(b^2-4ac)]/(2a)
If the equation of the parabola isy = ax^2 + bx + c then the roots are [-b +/- sqrt(b^2-4ac)]/(2a)
Set y = 0 and solve for x, with a parabola you should get one, two, or no x-axis crossings, it depends on the equation and the location on the x-y axis of the parabola.