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It is called the directrix.

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Q: What is the name of a line that you use to define a parabola?
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What is the name of the point you use to define a parabola?

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.


What is the name of the point that use to define a parabola?

i think its the vertex.


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


Do you use a line plot to order a data set?

no,you use a data when doing a vote. a line plot is for fractions or time line.


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.

Related questions

What is the name of the line that you use to define a parabola?

directrix


What is the name of the point you use to define a parabola?

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.


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 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.


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.


What is a parabola and quadratic?

A parabola is a line with one curve, that usually crosses the x-axis of a graph twice (unless the roots are imaginary). To find the roots, set y to zero and use the quadratic formula (-b±√b^2-4AC/2A)


Projectile motion with the use of parabola?

Yes.


A line that forms an edge of a figure?

The line that forms the edge of a figure is an edge line. Which forms the figure the has been drawn. You use this line to define a drawing or a figure.


Area bound by the straight line y equals x and the parabola y equals x2?

1st we find the intersection point of the straight line y=x & y=x^2, which are (0,0) & (1,1). after that we can use multiple integral with the 1st limit of y co-ordinate 0 to 1 & x limits x=y to x=square root(x). or simply we 1st calculate the area between the parabola & x-axis & the area between the straight line & x-axis in the interval 0 to 1. & after that we subtract the area of parabola by area of straight line.


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.


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.