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What is the name of the line that you use to define a parabola?
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∙ 14y ago
directrix
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∙ 14y ago
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Can you determine whether the inverse of a function is a function by using vertical line test?
Not quite. You can use a vertical line test on the graph of the inverse mapping, OR you can use a horizontal line test on the original graph. The horizontal line test is used in the same way.
Is x cubed minus 3 a one to one function?
use the horizontal line text, a horizontal line intersects the graph of x3 -3 only once so it is one to one.
Define The LIMIT as in Calculus?
The term "limit" in calculus describes what is occurring as a line approaches a specific point from either the left or right hand side. Some limits approach infinity while some approach specific points depending on the function given. If the function is a piece-wise function, the limit may not reach a specific value depending on the function given. For a more in-depth definition here is a good link to use: * http://www.math.hmc.edu/calculus/tutorials/limits/
Identify the vertex of the parabola y equals 4x2 - 24x plus 38?
There are two ways to solve this problem; one that involves basic calculus skills and one that can be done with algebraic methods. I will do both, and you can choose which one you understand based on your knowledge of mathematics. I will start with the algebraic method. The end goal of this is to get the equation in vertex form, where you will be able to simply read the vertex off of the equation. To do this, we will employ a method called "completing the square". Your original function is: y=4x2-24x+38 This is a parabola, but unfortunately it is not in vertex form. A parabola in vertex form will read as y=(x-a)2+c where "a" and "c" are real numbers. To get the component (x-a)2, we must have a trinomial that can be factored to be a product of the same term squared, that is it must be a trinomial that results from multiplying (x-a)*(x-a). In its current form, the trinomial in your equation is not a perfect square. You cannot express 4x2-24x+38 as product of two of the same binomial terms. So, we must make it a trinomial that is a perfect square. We will use the "completing the square" method. First, however, we must make it so no coefficient exists in front of the x2 term. This will make "completing the square" much easier. To do this, we will have to divide everything by 4. Remember what you do to one side of an equation you must do to both sides. y/4=4x2/4-24x/4+38/4 Simplify this to get: y/4=x2-6x+(19/2) To "complete the square", take the coefficent of the single-x term and halve it, then square it. For this trinomial, the single-x term is -6x: -6/2=3 32=9 This is the number that must be at the end of your trinomial to make it a perfect square. If you can manipulate the equation to get a trinomial of x2-6x+9, this will be a perfect square. Currently, you have x2-6x+(19/2). 19/2 is the same as 9.5. You could make 19/2 into the 9 that you need if you subtracted 1/2. You can do this as long as you do it to both sides of the equation. So, you get: y/4-(1/2)=x2-6x+(19/2)-(1/2) y/4-(1/2)=x2-6x+9 The trinomial you now have on the right side of the equation can be rewritten as (x-3)2, so you get: y/4-(1/2)=(x-3)2 By isolating y, you get an equation of a parabola in vertex form: y/4=(x-3)2+(1/2) y=4((x-3)3+(1/2)) y=4(x-3)3+2 This is vertex form, which reads as y=a(x-b)2+c. In this case, a=4, b=3, and c=2. You can get the vertex of a parabola in this form to be simply (b,c). So, the vertex for this parabola is (3,2). As a side note: the "a" value of a parabola in vertex form (4 for your equation) does not affect its vertex, it simply affects how "fat" or "skinny" the parabola will be. Second side note: If the squared term had happened to have been (x+3)2, the x coordinate of the vertex would be -3, not 3. Since vertex form reads as (x-a)2, "a" must be negative for an equation have (x+a)2 Now, for the calculus approach. This method is much simpler, but you must understand the basic of calculus to do it. I will not explain all of that, because it would take far too long. For this example, I will assume you know the meanings and applications of function maxima and minima are and how to derive a function. Since the function is a parabola, it will only have one critical point, its vertex, and since it is a parabola that opens upward, this critical point will be a minimum. So, the absolute minimum of the parabola is its vertex. By finding the absolute minimum, we find the vertex. To do this, we must first find the first derivative of the function. y=4x2-24x+38 y'=8x-24 By finding where the first derivative equals zero, we find the critical points to investigate on this function to determine if they are minima or maxima. y'=0 8x-24=0 8x=24 x=3 Since there is only one critical point, and since the function is a parabola, we know that this one point is the vertex. There is no need to verify that it is a minimum, although you can do that if you want to. We have found the x-value for the vertex of the function, so now we must find its corresponding y-value for the original function. By plugging in 3 for x in the original function, we get: y=4(3)2-24(3)+38 y=4*9-72=38 y=36-72+38 y=2 So, for x=3, y=2, meaning our vertex is (3,2).
What is a solution to the equation 5x 2y -1?
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals" etc. Also use ^ to indicate powers (eg x-squared = x^2). However, the question looks as if it might be a linear equation: an equation of a straight line. If that is the case then there are infinitely many solutions. The coordinates of each point on the line represents a solution.
What is the name of a line that you use to define a parabola?
It is called the 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.
