Write the equation of the parabola y = x^2 – 4x – 29 in standard form.?
1. Write the standard form of the equation for the circle that passes through the points (30, –2), (–1, –19), and (–18, 12). Then identify the center and radius. Choices: A. (x + 6)^2 + (y – 5)^2 = 625 (6, 5), 25 B. (x – 6)^2 + (y – 5)^2 = 625 (6, 5), 25 C. (x + 5)^2 – (y + 6)^2 = 650 (3, 5), 15 D. (x + 5)^2 + (y + 6)^2 = 625 (6, 2), 25 2. Write the equation in standard form of the ellipse with foci (8, 0) and (–8, 0) if the minor axis has y–intercepts of 2 and –2. Choices: A.(x – 68)^2+ (y – 4)^2= 1 B. (x+68)^2 -(y+4)^2 = 1 C.x^2/68 + y^2/4 = 1 D. x^2/32 - y^2/4 = 1 3. Write the equation of the parabola y = x^2 - 4x - 29 in standard form. A. y + 33 = (x – 2)^2 B. y + 29 = (x – 2)^2 C. y + 4 = (x – 2)^2 D. y – 2 = (x – 2)^2 Please show steps Thanks :) Will do best answer!
Write the new equation of the parabola y=( x - 3)^2 -7 that is translated 4 units left, 5 units up?
When you change horizontal stuff, you add it in the parentheses and you do the opposite of what you normally do (i.e. left is adding and right is subtracting). When you change horizontal stuff, you add it outside the parentheses and "what you see is what you get" (i.e. up is adding and down is subtracting). y = (x - 3)^2 - 7 Translate left by adding 4 to x - 3 y = (x - 3 + 4)^2 - 7 y = (x + 1)^2 - 7 Translate up by adding 5 to the end y = (x + 1)^2 - 7 + 5 y = (x + 1)^2 - 2 For the compression, I don't know whether you are trying to vertically compress or horizontally compress: If you're vertically compressing: y = ¼(x + 1)^2 - 2 If you're horizontally compressing: y = (4x + 4)^2 - 2
Write equation of parabola?
you need to do the method known as completing the square. y = x^2 -6x -1 y = (x^2 -6x +9 - 9) -1 y = (x^2 -6x +9) -10 y = (x - 3)^2 -10 Steps: 1. If the value in front of 'x' isn't 1 then factor it out of the first two terms 2. Divide the second term by 2 and then square it 3. Add and then subtract this value 4. Take the fourth thing in the brackets out 5. Combine like terms and factor
How to write parabola equations...?
Absolutely. First let's write out all the different equations associated with parabolas, then we can choose which we need based on the information given. Parabola's opening up or down (U or ∩): y = A(x - h)² + k A is the constant which determines the shape of the parabola. If A is positive the parabola opens up, if it is negative it opens down. The point (h,k) is the vertex. Parabolas opening left or right (⊂ or ⊃) x = B(y - k)² + h If B is negative it opens left, positive it opens right. It should be noted that these parabolas are not functions as they don't pass the vertical line test. Alright, with all that let's get the information we know. The vertex is at (5,-3), so we have h = 5, k = -3. This was the easy part, it gets a bit harder from here. Okay, so the focus is inside parabola. So because the focus is 3 units to the right of the vertex we're going to use the second equation and B will be positive. To find B we have to use the definition of a parabola, that is that it is the set of points that is equidistant from a point (the focus) and a line (the directrix). Because we have the focus and vertex, which we know are 3 units apart, the directrix is easy to find. It's 3 units the other way, and because the parabola is opening to the right, it is a vertical line. The directrix is x = 2. Now, to find B, we need another point on the parabola, any point other than the vertex will do. So we're going to find the point on the parabola directly above the focus, this is the easiest to find (or the one below the focus). So we know the x coordinate of this point is 8, and the distance to the directrix is 6. So we need the distance to the focus to be 6 as well, but since they're on the same vertical line this boils down the distance between the y-coordinates. So our point is (8,3). Plug this point and the vertex into the formula for the parabola to solve for B. 8 = B(3 - (-3))² + 5 3 = B6² 3/36 = 1/12 = B So our parabola is x = (1/12)(y + 3)² + 5. It should be noted that there are formulas to find A and B, but they're based on the same exact process I walked you through, just in general. Try doing this process in general if you're interested in those formulae. I hope this has helped. Good luck, and don't forget pictures are your friend. Have a nice day!
Writing the equation of a parabola?
Vertex form: y = a * (x - h)^2 + k (h , k) is the vertex y = a * (x - (-2))^2 - 8 y = a * (x + 2)^2 - 8 y = -494 , x = 7 -494 = a * (7 + 2)^2 - 8 -494 + 8 = a * 9^2 -486 = a * 81 a = -6 y = -6 * (x + 2)^2 - 8 f(x) = 5x^2 is in vertex form already: f(x) = 5 * (x - 0)^2 + 0 All we have to do is change our vertex f(x) = 5 * (x - (-13))^2 + (-16) f(x) = 5 * (x + 13)^2 - 16
What is the equation of a parabola?
[math]a_{11}x^2+2a_{12}xy+a_{22}y^2+b_1x+b_2y+c=0[/math]is the general equation of a quadratic curve if [math]|a_{11}+|a_{12}|+|a_{22} |>0[/math]. Its nature is determined by the discriminant [math]D=a_{11}a_{22}-a_{12}^2.[/math] If[math]D>0[/math] it is an ellipse unless [math]a_{11}=a_{22}\ \mathrm{and}\ a_{12}=0,[/math] then it is a circle.[math]D=0[/math] it is a parabola[math]D<0[/math] it is a hyperbola.You can get rid of the [math]xy[/math] term by a rotation:[math]x'=x\cos\phi+y\sin\phi,\ y'=-x\sin\phi+y\cos\phi[/math]and choose [math]\phi[/math] cleverly.You can get rid of the [math]x[/math] and [math]y[/math] terms by a translation[math]x''=x'+d_1,\ y''=y'+d_2.[/math]You may also write the equation in vector form, [math]\mathbf x=(x,y)^T[/math]:[math]\mathbf x^TA\mathbf x+\mathbf b^T\mathbf x + c=0[/math]with [math]A[/math] a symmetric [math]2\times2[/math] matrix. If the eigenvalues of [math]A[/math] have equal sign it is an ellipse (circle if equal), if one of them is 0 it is a parabola, if they have opposite sign it is a hyperbola. Since [math]A[/math] is symmetric it is diagonizable: [math] A=C^T\Lambda C,[/math] which gives you the rotation matrix [math]C[/math] straight away.
Given the equation y=-2x^2, write an equation of a parabola if the graph has been widened?
The “parent” graph for a quadratic in vertex form is y = x^2, then if there are any transformations the graph is written as y = a(x - h)^2 + k.This graph is y = -2x^2. So a is -2. This means the parabola has been reflected across the x-axis (it is upside down) and it has been dilated (or stretched) by a factor of 2 in the y direction.You could rewrite this as y = - (sqrt(2) x)^2. Now your transformation is a transformation of x instead of a transformation of y. A transformation of x is always the opposite of what you see. So you would say this parabola is 1/sqrt(2) times as wide as the parent graph. That means it is actually skinnier not wider.
How do you write the equation of the parabola in vertex form?
The equation of parabola is given by :y^2 = 4 a x {If the axis of parabola is x-axis}…(i)and;x^2 = 4 a y {If the axis of parabola is y-axis}…(ii)in both the above equations {i.e, eq(i) and eq(i)}the vertex is Origin i.e, (0,0).if the vertex is (h,k) then the above written equations become:(y-k)^2 = 4 a (x-h) {If the axis of parabola is x-axis}…(iii)and(x-h)^2 = 4 a (y-k) {If the axis of parabola is y-axis}…(iv).So equation (iii) and equation (iv) are the required equations. Ans.I think it is a little bit difficult way . But if you try to realize that “what do the literals (alphabets) in this equation means and how the distance formula is used”, then it will be very easy for you to remember these equations.you can ask me if you don’t know about the literals of them.Thanks!!!
Math Please help! Write a equation of a parabola?
In the general form y = a*(x - h)^2 + k, the vertex is (h,k) and the value a determines whether the parabola opens upward or downward and whether it is 'squeezed' or 'stretched'. A) In this case we have (h,k) = (6,2) and a = -3, since 'congruence' means that the parabolas share the same value for a. So the equation of this parabola is y = -3*(x - 6)^2 + 2. B) The axis of symmetry indicates the x-coordinate of the vertex, the maximum value indicates the y-coordinate of the vertex and congruence to y = (2/3)*x^2 gives us the value for a. So (h,k) = (5,-3) and a = (2/3), and so the equation is y = (2/3)*(x - 5)^2 - 3. C) In this case we have the vertex but not a. To find a we need to plug A(-1,-4) into the equation y = a*(x - (-3))^2 + 4 = a*(x + 3)^2 + 4: -4 = a*(-1 + 3)^2 + 4 ----> -8 = a*2^2 ----> -8/4 = a ----> a = -2. So the equation of the parabola is y = -2*(x + 3)^2 + 4.
How do you write the equation of a parabola given the vertex and y-intercept?
Let's start with the answer. The vertex is the lowest point when ax^2 opens upward. Another way to say that is that a>0. y = (x - 2)^2 is where this happens. y = a(x - 2)^2 + 0 is the final answer. which shortens down to y = a(x - 2)^2 The y intercept tells you that a = 1. Here's how. y = a(x - 2) ^2 The y intercept occurs when x = 0 y = a(0 - 2)^2 4 = a(-2)^2 4 = a*4 a = 1. Now suppose that you had 2,0 and 0,8 instead. y = a(x - 2)^2 + 0 just as before, but the 8 has to be considered. 8 = a(x - 2)^2 8 = a (0 - 2)^2 8 = a*4 a = 2 answer y = 2(x - 2)^2 One more example. Suppose the vertex is (2 ,-4) and the y intercept is 0,6 Now what? y = a(x - 2)^2 - 4 Notice what happened. When 2 is put in for x, the result is - 4. That's the y value of - 4 which is as small as it gets. Now when x = 0, we find out what a is y = a(0 - 4)^2 - 4 6 = a(16) - 4 Add 4 to both sides. 10 = a(16) a = 10/16 = 5/8 It isn't pretty, but that's the method. Always do this question in the same order b,0 means a(x -b)^2 0,c means x = 0 and you solve for a.