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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.
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What is the name of the point that use to define a parabola?
i think its the vertex.
What is the name of a line that you use to define a parabola?
It is called the directrix.
Find the area S of the region bounded by the parabola y 5x2 the tangent line to this parabola at 2 20 and the x-axis?
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
How do you graph quadratic functions in vertex form?
The standard form of quadratic function is: f(x) = a(x - h)^2 + k, a is different than 0 The graph of f is a parabola whose vertex it is the point (h, k). If a > 0, the parabola opens upward; if a < 0, the parabola opens downward. Furthermore, if |a| is small, the parabola opens more flatly than if |a| is large. It is a general procedure for graphing parabolas whose equations are in standard form: Example 1: Graph the the quadratic function f(x) = -2(x - 3)^2 + 8 Solution: Standard form: f(x) = a(x - h)^2 + k Given function: f(x) = -2(x - 3) + 8 From the give function we have: a= -2; h= 3; k = 8 Step 1. Determine how the parabola opens. Note that a = -2. Since a < 0, the parabola is open downward. Step 2. Find the vertex. The vertex of parabola is at (h, k). because h = 3 and k = 8, the parabola has its vertex at (3, 8). Step 3. Find the x-intercepts by solving f(x) = 0. Replace f(x) with 0 at f(x) = -2(x - 3)^2 + 8 and solve for x 0 = -2(x - 3)^2 + 8 2(x - 3)^2 = 8 (x- 3)^2 = 4 x - 3 = square radical 4 x - 3 = 2 or x -3 = -2 x = 5 or x = 1 The x- intercepts are 1 and 5. Thus the parabola passes through the points (1, 0) and (5, 0), this means that parabola intercepts the x-axis at 1 and 5. Step 4. Find the y-intercept by computing f(0). Replace x with 0 in f(x) = _2(x - 3)^2 + 8 f(0) = -2(0 - 3)^2 + 8 f(0) = -2(9) + 8 f(0) = -10 The y-intercept is -10. Thus the parabola passes through the point (0, -10), this means that parabola intercepts the y-axis at -10. Step 5. Graph the parabola. With a vertex at (3, 8), x-intercepts at 1 and 5, and a y-intercept at -10. The axis of symmetry is the vertical line whose equation is x = 3. Example 2: Graphing a quadratic function in the form f(x) = ax^2 + bx + c Graph the quadratic function f(x) = -x^2 - 2x + 1 Solution: Here a = -1, b = -2, and c = 1 Step 1. Determine how the parabola opens. Since a = 1, a < 0, the parabola opens downward. Step 2. Find the vertex. We know that x-coordinate of the vertex is x = -b/2a. Substitute a with -1 and b with -2 into the equation for the x-coordinate: x = - b/2a x= -(-2)/(2)(-1) x = -1, so the x-coordinate of the vertex is -1, and the y-coordinate of the vertex will be f(-1). thus the vertex is at ( -1, f(-1) ) f(x) = -x^2 - 2x +1 f(-1) = -(-1)^2 - 2(-1) + 1 f(-1) = -1 + 2 + 1 f(-1) = 2 So the vertex of the parabola is (-1, 2) Step 3. Find the x-intercepts by solving f(x) = o f(x) = -x^2 -2x + 1 0 = -x^2- 2x + 1 We can't solve this equation by factoring, so we use the quadratic formula to solve it. we get to solution: One solution is x = -2.4 and the other solution is 0.4 (approximately). Thus the x-intercepts are approximately -2.4 and 0.4. The parabola passes through ( -2.4, 0) and (0.4, 0) Step 4. Find the y-intercept by computing f(0). f(x) = -x^2 - 2x + 1 f(0) = -(0)^2 - 2(0) + 1 f(0) = 1 The y-intercept is 1. The parabola passes through (0, 1). Step 5. graph the parabola with vertex at (-1, 2), x-intercepts approximately at -2.4 and 0.4, and y -intercept at 1. The line of symmetry is the vertical line with equation x= -1.
How do you use over as a preposition?
Use the word with a noun to define what it is over, or about. "He jumped over the fence." "The key is over the door." "They argued over the new tax plan."
What is the name of the point that use to define a parabola?
i think its the vertex.
What is the name of the point that you use to define a parabola?
it's called the focus
What is the name of the line that you use to define a parabola?
directrix
What is the name of a line that you use to define a parabola?
It is called the directrix.
Is there another not as common name for a 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.
Projectile motion with the use of parabola?
Yes.
An equilateral triangle is inscribed in a parabola with its vertex at the vertex of the parabola how do you find the length of the equilateral triangle?
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.
How do you do a parabola?
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.
What careers use parabolas?
One career that might use a parabola is a mathematics teacher. Geometry teachers might also use parabolas. A parabola is a line consisting of points that are connected and spaced unilaterally.
How did Rene Descartes use the parabola?
Descartes used the parabola to illustrate algebraic equations. He put these equations on a visible plane using the Cartesian coordinate system and they sometimes took the shape of a "u" curve, or a parabola.
Use the word parabola in a sentence?
A parabola is the curved path of a projectile is the result of constant motion horizontallt and accelerated motion vertically under the influrnce of gravity.
How do you find the equation of a parabola if you know the vertex and a point it passes through?
Use this form: y= a(x-h)² + k ; plug in the x and y coordinates of the vertex into (h,k) and then the other point coordinates into (x,y) and solve for a.
