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To find the vertex of a parabola given its equation in standard form (y = ax^2 + bx + c), you can use the formula for the x-coordinate of the vertex: (x = -\frac{b}{2a}). Once you have the x-coordinate, substitute it back into the equation to find the corresponding y-coordinate. Thus, the vertex can be expressed as the point ((-\frac{b}{2a}, f(-\frac{b}{2a}))). For parabolas in vertex form (y = a(x-h)^2 + k), the vertex is simply the point ((h, k)).
What else can I help you with?
How do you find the endpoints of a parabola when you have the function of the line?
A parabola has no endpoints: it extends to infinity.A parabola has no endpoints: it extends to infinity.A parabola has no endpoints: it extends to infinity.A parabola has no endpoints: it extends to infinity.
How do you draw a parabola on an equilateral triangle?
To graph a parabola you must find the axis of symmetry, determine the focal distance and write the focal as a point, and find the directrix. These are all the main points you need to be able to draw a parabola.
How do you find a parabola opening up or down?
If you can mash the equation for the parabola into the form Y = Ax2 + Bx + C, then the parabola opens up if 'A' is positive, and down if 'A' is negative.
To find the value of a in a parabola opening left or right subtract the x value of the parabola at the vertex from the x value of the point on the parabola that is one unit the vertex?
Above
To find the value of a in a parabola opening up or down, subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the of the vertex?
right
How do you find the endpoints of a parabola when you have the function of the line?
A parabola has no endpoints: it extends to infinity.A parabola has no endpoints: it extends to infinity.A parabola has no endpoints: it extends to infinity.A parabola has no endpoints: it extends to infinity.
To find the value of a in a parabola opening up or down subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the of the vertex?
To find the value of a in a parabola opening up or down subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the right of the vertex.
How do you find the equation of a parabola if you know the equation of the tangent that touches it?
You need more than one tangent to find the equation of a parabola.
How do you draw a parabola on an equilateral triangle?
To graph a parabola you must find the axis of symmetry, determine the focal distance and write the focal as a point, and find the directrix. These are all the main points you need to be able to draw a parabola.
How do you find a parabola opening up or down?
If you can mash the equation for the parabola into the form Y = Ax2 + Bx + C, then the parabola opens up if 'A' is positive, and down if 'A' is negative.
To find the value of a in a parabola opening up or down, subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the of the vertex?
right
To find the value of a in a parabola opening left or right subtract the x value of the parabola at the vertex from the x value of the point on the parabola that is one unit the vertex?
Above
How do you find the gradient of a parabola?
(-1.5,0) (1.5,0) what is the gradient?
How do you find points on a parabola to graph them?
Select a set of x values and find the value of y or f(x) - depending on how the parabola is defined. These are the values that you need to graph.
How do you find the zeros in a parabola?
If the equation of the parabola isy = ax^2 + bx + c then the roots are [-b +/- sqrt(b^2-4ac)]/(2a)
What are the roots of a parabola?
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.
How do you find centroid of parabola?
The centroid of a parabola is found with the equation y = h/b^2 * x^2, where the line y = h. Additionally, the area is 4bh/3.
