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.
No you can't. There is no unique solution for 'x' and 'y'. The equation describes a parabola, and every point on the parabola satisfies the equation.
If you know the equation, you just plug in x = 0 and solve.
If the equation of the parabola isy = ax^2 + bx + c then the roots are [-b +/- sqrt(b^2-4ac)]/(2a)
The focus of a parabola is a fixed point that lies on the axis of the parabola "p" units from the vertex. It can be found by the parabola equations in standard form: (x-h)^2=4p(y-k) or (y-k)^2=4p(x-h) depending on the shape of the parabola. The vertex is defined by (h,k). Solve for p and count that many units from the vertex in the direction away from the directrix. (your focus should be inside the curve of your parabola)
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.
Once you calculate the X coordinate using the axis of symmetry (X=-b/2a), you plug that value in for all of the X's in the equation of the parabola. You then solve the equation for the value of Y.
You don't really solve the equation. You use it. Having said that, see the Wikipedia article, which has an adequate discussion of the equation and shows it in a few forms.
The formula for the Latus rectum is simply 2L = 4a with a stands for the distance of the focus from the vertex of the parabola. Given a, you can simply solve for the length of the latus rectum by using this formula.. L = 2a
y=b+x+x^2 This is a quadratic equation. The graph is a parabola. The quadratic equation formula or factoring can be used to solve this.
Y=a(x-h)+k is the vertex formula. Since the vertex is at (-2,-3) this parabola has the equation: y=a(x+2)^2-3 We can plug in x=-1 but we really need to know a, to solve for y. ( we can solve it, but we will have an a in the solution)
First you need more details about the parabola. Then - if the parabola opens upward - you can assume that the lowest point of the triangle is at the vertex; write an equation for each of the lines in the equilateral triangle. These lines will slope upwards (or downwards) at an angle of 60°; you must convert that to a slope (using the tangent function). Once you have the equation of the lines and the parabola, solve them simultaneously to check at what points they cross. Finally you can use the Pythagorean Theorem to calculate the length.
the equation of a parabola is: y = a(x-h)^2 + k *h and k are the x and y intercepts of the vertex respectively * x and y are the coordinates of a known point the curve passes though * solve for a, then plug that a value back into the equation of the parabola with out the coordinates of the known point so the equation of the curve with the vertex at (0,3) passing through the point (9,0) would be.. 0 = a (9-0)^2 + 3 = 0 = a (81) + 3 = -3/81 = a so the equation for the curve would be y = -(3/81)x^2 + 3
you don't answer an equation, you solve an equation
Sure. You can always 'solve for' a variable, and if it happens to be the only variable in the equation, than that's how you solve the equation.
Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a â‰ 0. The roots of the parabola are given by x = [-b Â± 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.
One useful strategy is having started a question to complete it.
A calculator can be used to proportions to answer a equation. This is easier to solve when having variables on both sides.
If you solve such an equation for "y", you get an equation in the slope-intercept form.
you can only solve for one in an equation so it can equal something
It is not an equation if it does not have an equals sign. You could simplify it but not solve it.
How do you use division to solve a multiplication equation?Answer this question…
Solution 1Start by putting the parabola's equation into the form y = ax2 + bx + c if it opens up or down,or x = ay2 + by + c if it is opens to the left or right,where a, b, and c are constants.The x-value for the vertex is -(b/2a). You can use this x-value to solve for the y-value by substituting the x value in the original quadratic equation.Solution 2Put the parabola's equation into this form: y - k = 4p(x - h)2or x - h = 4p(y - k)2You just need to simplify the equation until it looks like this. The vertex is located at the coordinates (h,k). (p is for the focus, but that isn't important as long as you know h and k.)
You solve the equation.