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.
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Suppose the equation of the parabola is y = ax2 + bx + c Now, where the parabola crosses the x-axis (the x intercepts), the value of y must be zero (that is what crossing the x-axis means). If the discriminant, b2 - 4ac is less than zero, y has no real roots. This means that there is no real value of x for which y equals zero and so the parabola has no x intercepts. If the discriminant is zero then the parabola only touches the x-axis - at (-b/2a,0). If the discriminant is greater than zero, there are two distinct intercepts. If a>0 then the parabola is shaped like a U and is wholly above the x-axis. If a<0 then the parabola is an upturned U, wholly below the x axis. If a = 0 the quadratic term disappears and the function is a straight line, not a parabola.
To have a parabola with only one x-intercept, the vertex of the parabola must lie on the x-axis. This means the parabola opens either upwards or downwards, depending on the coefficient of the squared term in the equation. If the coefficient is positive, the parabola opens upwards, and if it is negative, the parabola opens downwards. By adjusting the coefficients in the equation of the parabola, you can position the vertex such that there is only one x-intercept.
NONE If both roots are imaginary, the means the parabola does NOT cross the x-axis at all. The place where a function crosses the x- axis has the coordinate (x,0) for some value of x. That means if you plug in x to the function or polynomial, you get 0. This is equivalent to saying that x is a root of the polynomial. But if the only roots are imaginary, there will be no point (x,0) for any real number x.
Look at the discriminant, B2 - 4AC, in the quadratic equation. As it goes from negative to positive, the parabola moves in the direction of its small end towards the X-axis. At zero, it touches the X-axis.