Let us say you X intercepts are -2 and 3 set up (X + 2)(X - 3) FOIL X^2 - X - 6 = 0 ----------------------- your parabolic equation
Factorise equation, and look at what x values are needed for the equation to equal zero. Eg. x^2+5x+6 (x+3)(x+2)=0 So parabola intercepts x axis at -3 and -2.
If the discriminant is negative, there are 0 interceptsIf the discriminant is zero, there is 1 interceptIf the discriminant is positive, there are 2 intercepts
The zeros of functions are the solutions of the functions when finding where a parabola intercepts the x-axis, hence the other names: roots and x-intercepts.
No, but any parabola can be transformed into the form y = x^2.
the vertex of a parabola is the 2 x-intercepts times-ed and then divided by two (if there is only 1 x-intercept then that is the vertex)
Let us say you X intercepts are -2 and 3 set up (X + 2)(X - 3) FOIL X^2 - X - 6 = 0 ----------------------- your parabolic equation
x^2+8x=20 x^2+8x-20=0 (x+10)(x-2)=0 x=-10 and x=2 are the roots (intercepts) (-10,0) and (2,0) are the x-intercepts.
Factorise equation, and look at what x values are needed for the equation to equal zero. Eg. x^2+5x+6 (x+3)(x+2)=0 So parabola intercepts x axis at -3 and -2.
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.
If the discriminant is negative, there are 0 interceptsIf the discriminant is zero, there is 1 interceptIf the discriminant is positive, there are 2 intercepts
The zeros of functions are the solutions of the functions when finding where a parabola intercepts the x-axis, hence the other names: roots and x-intercepts.
No, but any parabola can be transformed into the form y = x^2.
there are three main characteristics of a parabola. these are: 1. vertex: the point at the apex of a parabola 2. x- intercepts: the points at which the parabola intersects or touches the x axis. 3. face: if the parabola is in the form of the letter "u" then it's face is upwards. if the parabola is the in form of the inverted letter "u" then it face downwards :D
x=2,-6 x^2+4x-12 (x^2-2x)(+6x-12) x(x-2)+6(x-2) (x-2)(x+6)=0
Solve the quadratic equation.If y = ax^2+bx+c then the intercepts arex = [-b +/- sqrt(b^2 - 4ac)]/(2a).The solutions are real if and only if b^2 >= 4ac
No, a parabola does not have to have an x-intercept. ex. -2(x-2)^2 - 4 is a parabola that has no x-intercept.