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The standard equation for a Parabola with is vertex at the origin (0,0) is, x2 = 4cy if the parabola opens vertically upwards/downwards, or y2 = 4cx when the parabola opens sideways.
As the focus is at (0,6) then the focus is vertically above the vertex and we have an upward opening parabola.
Note that c is the distance from the vertex to the focus and in this case has a value of 6 (a positive number).
The equation is thus, x2 = 4*6y = 24y
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What is the coefficient of the squared term in the parabola's equation when the vertex is at 2 -1 and the point 5 0 is on it?
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (2, -1), and a point on it is (5, 0) → 0 = a(5 - 2)² + -1 → 0 = a(3)² -1 → 1 = 9a → a = 1/9 → The coefficient of the x² term is 1/9
How do you solve quadratics using graphing?
Recall that the graph of a linear equation in two variables is a line. The equation y = ax^2 + bx + c, where a, b, and c are real numbers and a is different than 0 represents a quadratic function. Its graph is a parabola, a smooth and symmetric U-shape. 1. The axis of symmetry is the line that divides the parabola into two matching parts. Its equation is x = -b/2a 2. The highest or lowest point on a parabola is called the vertex (also called a turning point). Its x-coordinate is the value of -b/2a. If a > 0, the parabola opens upward, and the vertex is the lowest point on the parabola. The y-coordinate of the vertex is the minimum value of the function. If a < 0, the parabola opens downward, and the vertex is the highest point on the parabola. The y-coordinate of the vertex is the maximum value of the function. 3. The x-intercepts of the graph of y = ax^2 + bx + c are the real solutions to ax^2 + bx + c = 0. The nature of the roots of a quadratic function can be determined by looking at its graph. If you see that there are two x-intercepts on the graph of the equation, then the equation has two real roots. If you see that there is one x-intercept on the graph of the equation, then the equation has one real roots. If you see that the graph of the equation never crosses the x-axis, then the equation has no real roots. The roots can be used further to determine the factors of the equation, as (x - r1)(x -r2) = 0
The vertex of this parabola is at 3 -2 When the x value is 4 the y value is 3 What is the coefficient of the squared expression in the parabolas equation?
Vertex = (3, - 2)Put in vertex form.(X - 3)2 + 2X2 - 6X + 9 + 2 = 0X2 - 6X + 11 = 0=============The coefficeint of the squared term is 1. My TI-84 confirms the (4, 3) intercept of the parabola and the 11 Y intercept shown by the function.
What is the line of symmetry for the parabola whose equation is y -x2 x 3?
Did you mean a parabola with equation y=3x^2? The line of symmetry is x=0 or the y-axis.
Using the given equations of parabolas find the focus the directrix and the equation of the axis of symmetry of x2 -8y?
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What is the standard equation for vertex at origin opens down 1 and 76 units between the vertex and focus?
Since the vertex is at the origin and the parabola opens downward, the equation of the parabola is x2 = 4py, where p < 0, and the axis of symmetry is the y-axis. So the focus is at y-axis at (0, p) and the directrix equation is y = -p. Now, what do you mean with 1 and 76 units? 1.76 units? If the distance of the vertex and the focus is 1.76 units, then p = -1.76, thus 4p = -7.04, then the equation of the parabola is x2 = -7.04y.
What is the equation of a prabola with the vertex 0 0 and focus 0 4?
x2 = 16y The standard formula for a parabola with its vertex at the origin (0, 0) and a given focus (and the y-axis as an axis of symmetry) is as follows: x2 = 4cy In this case, the c is the y value of the focus. The focus in this case was (0, 4), and the y value in the focus is 4. That makes the c = 4. Further, that makes the equation for this parabola x2 = 4 (c)y = 4 (4)y = 16y Given that the vertex was the origin, (0, 0), and the focus is (0, 4), we can conclude that the axis of symmetry is the y-axis because the y value of the focus is 0. We can also conclude that the parabola opens up, because the focus has a positive y value.
How do you solve parabola equation having lb focus at 0 3?
There is not enough information. You need either the directrix or vertex (or some other item of information).
What does 4 stands for in equation of parabola square of square of y equals 4ax?
A parabola with an equation, y2 = 4ax has its vertex at the origin and opens to the right. It's not just the '4' that is important, it's '4a' that matters. This type of parabola has a directrix at x = -a, and a focus at (a, 0). By writing the equation as it is, the position of the directrix and focus are readily identifiable. For example, y2 = 2.4x doesn't say a great deal. Re-writing the equation of the parabola as y2 = 4*(0.6)x tells us immediately that the directrix is at x = -0.6 and the focus is at (0.6, 0)
The vertex of this parabola is at -3 -1 When the y-value is 0 the x-value is 4 What is the coefficient of the squared term in the parabolas equation?
The vertex of this parabola is at -3 -1 When the y-value is 0 the x-value is 4. The coefficient of the squared term in the parabolas equation is 7
What is the standard form of the equation of the parabola with vertex 00 and directrix y4?
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
What is the coefficient of the squared term in the parabola's equation when the vertex is at 2 -1 and the point 5 0 is on it?
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (2, -1), and a point on it is (5, 0) → 0 = a(5 - 2)² + -1 → 0 = a(3)² -1 → 1 = 9a → a = 1/9 → The coefficient of the x² term is 1/9
Find equation what parabola its vertex is 0 0 and it passes through point 2 12 express the equation in standard form?
Y=3x^2 and this is in standard form. The vertex form of a prabola is y= a(x-h)2+k The vertex is at (0,0) so we have y=a(x)^2 it goes throug (2,12) so 12=a(2^2)=4a and a=3. Now the parabola is y=3x^2. Check this: It has vertex at (0,0) and the point (2,12) is on the parabola since 12=3x2^2
What is the parabola equation?
the equation of a parabola is: y = a(x-h)^2 + k *h and k are the x and y intercepts of the vertex respectively * x and y are the coordinates of a known point the curve passes though * solve for a, then plug that a value back into the equation of the parabola with out the coordinates of the known point so the equation of the curve with the vertex at (0,3) passing through the point (9,0) would be.. 0 = a (9-0)^2 + 3 = 0 = a (81) + 3 = -3/81 = a so the equation for the curve would be y = -(3/81)x^2 + 3
What is the coefficient of the x² term of the parabola with a vertex at -3 -1 and passes through the point 4 0?
A parabola with vertex (h, k) has equation: y = a(x - h)² + k With vertex (-3, -1) this becomes: y = a(x - -3)² + -1 = a(x + 3)² - 1 The point (4, 0) is on this parabola so: 0 = a(4 + 3)² - 1 → 7²a = 1 → a = 1/49 Thus the coefficient of x² is 1/49.
How are the vertices of the parabolas related to the equation of the quadratic function?
Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a ≠0. The roots of the parabola are given by x = [-b ± sqrt(D)]/2a where D is the discriminant. Rather than solve explicitly for the coordinates of the vertex, note that the vertical line through the vertex is an axis of symmetry for the parabola. The two roots are symmetrical about x = -b/2a so, whatever the value of D and whether or not the parabola has real roots, the x coordinate of the vertex is -b/2a. It is simplest to substitute this value for x in the equation of the parabola to find the y-coordinate of the vertex, which is c - b2/2a.
An equation of a parabola that has x equals 2 as its axis of symmetry is?
How about y = (x - 2)2 = x2 - 4x + 4 ? That is the equation of a parabola whose axis of symmetry is the vertical line, x = 2. Its vertex is located at the point (2, 0).
