All of the points on a parabola define a parabola.
However, the vertex is the point in which the y value is only used for one point on the parabola.
i think its the vertex.
It is called the directrix.
To determine the equation of a parabola with a vertex at the point (5, -3), we can use the vertex form of a parabola's equation: (y = a(x - h)^2 + k), where (h, k) is the vertex. Substituting in the vertex coordinates, we have (y = a(x - 5)^2 - 3). The value of "a" will determine the direction and width of the parabola, but any equation in this form with varying "a" values could represent the parabola.
The name of a point that is at (0,0) on a coordinate plane is called the origin.
First we need to find the equation of the tangent line to the parabola at (2, 20).Step 1. Take the derivative of the function of the parabola.Let f(x) = 5x^2f'(x) = 10xStep 2. Find the slope of the tangent line at x = 2. Evaluate f'(2).f'(2) = 2 x 10 = 20Step 3. Using the slope, m = 20, and the point (2, 20), find the equation of the tangent line at that point. Use the point-slope form of a line(y - y1) = m(x - x1)(y - 20) = 20(x - 2)y - 20 = 20x - 40 add 20 to both sidesy = 20x - 20Step 4. Find the points of intersections of y = 5x^2 and y = 20x - 205x^2 = 20x - 20 Divide by 5 to both sidesx^2 = 4x - 4 subtract 4x and add 4 to both sidesx^2 - 4x + 4 = 0 factor(x - 2)^2= 0x = 2Step 5. Find the intersection of the tangent line with x-axis.y = 20x - 20y = 020x - 20 = 0x = 1Since the vertex of the parabola is (0, 0) and the intersection of the tangent line with parabola is (2,20) we use the interval [0, 2] to fin the required area.Step 6. IntegrateA = ∫ [(5x^2)] dx, where the below boundary is 0, and the upper boundary is 2 minus A= ∫ (20x + 20)] dx from 1 to 2= 10/3
i think its the vertex.
it's called the focus
directrix
It is called the directrix.
To find the "a" value in a parabola, which determines its width and direction (opening upwards or downwards), you can use the standard form of a quadratic equation: (y = ax^2 + bx + c). If you have a specific point on the parabola and the values of (b) and (c), you can substitute these into the equation along with the coordinates of the point to solve for (a). Alternatively, if the parabola is in vertex form, (y = a(x-h)^2 + k), you can derive (a) using the vertex and another point on the curve.
To determine the equation of a parabola with a vertex at the point (5, -3), we can use the vertex form of a parabola's equation: (y = a(x - h)^2 + k), where (h, k) is the vertex. Substituting in the vertex coordinates, we have (y = a(x - 5)^2 - 3). The value of "a" will determine the direction and width of the parabola, but any equation in this form with varying "a" values could represent the parabola.
The parabola is a type of conic section, . The problem is that this is not a descriptive as the if the word "parabola" is used. The reason is that it is not the only geometric shape that can be derived by slicing a cone with a plane. Use the link below to see a drawing and learn more.
To find the equation of a parabola with vertex at ((-3, 0)) that passes through the point ((3, 18)), we can use the vertex form of a parabola, (y = a(x + 3)^2). To determine the value of (a), substitute the point ((3, 18)) into the equation: [ 18 = a(3 + 3)^2 \implies 18 = a(6)^2 \implies 18 = 36a \implies a = \frac{1}{2}. ] Thus, the equation of the parabola is (y = \frac{1}{2}(x + 3)^2).
To write an equation for a parabola in standard form, use the format ( y = a(x - h)^2 + k ) for a vertical parabola or ( x = a(y - k)^2 + h ) for a horizontal parabola. Here, ((h, k)) represents the vertex of the parabola, and (a) determines the direction and width of the parabola. If (a > 0), the parabola opens upwards (or to the right), while (a < 0) indicates it opens downwards (or to the left). To find the specific values of (h), (k), and (a), you may need to use given points or the vertex of the parabola.
Yes.
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.
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