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How do u find the equation of the axis of symmetry and the vertex of the graph of each function for example y x2-8x-9 Plz help i need to know this?
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What is the vertex and line of symmetry of the parabola when y equals x2 -8x plus 1 taking pi as 3.142 showing work?
Forget about thw value of pi because it has nothing to do with the question. y = x2-8x+1 can be expressed as y = (x-4)2-15 Information about the vertex and line of symmetry can be extracted from the rearranged equation such as:- Vertex: (4, -15) Line of symmetry: x = 4
Is x2 plus 8 a function?
x2+8= y This equation represents a function. It will be a parabola with the vertex at (0,8). You can easily graph this on a graphing calculator or from prior knowledge. You know the basic graph of y=x2 with vertex (0,0) and opens upwards on the y-axis. From the equation, you simply shift the vertex vertically up 8 so the new vertex is (0,8) This represents a function because for every x value there is one y value.
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.
How do you find the y-coordinate vertex of a parabola?
Once you calculate the X coordinate using the axis of symmetry (X=-b/2a), you plug that value in for all of the X's in the equation of the parabola. You then solve the equation for the value of Y.
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.
What is the equation of the axis of symmetry and the vertex of the graph gx-2x-12x 6?
There is no equation (nor inequality) in the question so there can be no graph - with or without an axis of symmetry.
What is the vertex and the line of symmetry for fx equals 5xsquared?
Vertex = (0,0) Line of symmetry = y axis You should of known that as this function is only X^2
Fill in the blank The of the vertex of a quadratic equation is determined by substituting the value of x from the axis of symmetry into the quadratic equation?
D
What things are significant about the vertex of a quadratic function?
It is a turning point. It lies on the axis of symmetry.
Find the equation of the axis symmetry and the coordinates of the vertex of the graph of each function for y equals 2x plus 4?
I'm assuming that you meant y = 2(x^2) +4. If it were only y = 2x +4, then this would be a linear equation and not a parabola. Anyways, use the equation x = -b/2a to find the x-value of your vertex AND your axis of symmetry. (Given the standard equation y = a(x^2) + bx + c) So, x = -0/2(2) - x = 0 (Axis of Symmetry) Now plug 0 back into your equation to find your y-value of your vertex. y = 2(0^2)+4 y=0 + 4 y = 4 Therefore Vertex = (0,4)
Example of lines of symmetry in a nonagon?
A line of symmetry must go from one vertex to the midpoint of the opposite side.
What different information do you get from vertex form and quadratic equation in standard form?
The graph of a quadratic function is always a parabola. If you put the equation (or function) into vertex form, you can read off the coordinates of the vertex, and you know the shape and orientation (up/down) of the parabola.
Find the vertex and equation of the directri for y2 equals -32x?
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
What are the vertex and the axis of symmetry of the equation y equals 2x² plus 4x - 10?
In the form y = ax² + bx + c the axis of symmetry is given by the line x = -b/2a The axis of symmetry runs through the vertex, and the vertex is given by (-b/2a, -b²/4a + c). For y = 2x² + 4x - 10: → axis of symmetry is x = -4/(2×2) = -4/4 = -1 → vertex = (-1, -4²/(4×2) - 10) = (-1, -16/8 - 10) = (-1, -12)
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).
What is the vertex and line of symmetry of the parabola when y equals x2 -8x plus 1 taking pi as 3.142 showing work?
Forget about thw value of pi because it has nothing to do with the question. y = x2-8x+1 can be expressed as y = (x-4)2-15 Information about the vertex and line of symmetry can be extracted from the rearranged equation such as:- Vertex: (4, -15) Line of symmetry: x = 4
Is x2 plus 8 a function?
x2+8= y This equation represents a function. It will be a parabola with the vertex at (0,8). You can easily graph this on a graphing calculator or from prior knowledge. You know the basic graph of y=x2 with vertex (0,0) and opens upwards on the y-axis. From the equation, you simply shift the vertex vertically up 8 so the new vertex is (0,8) This represents a function because for every x value there is one y value.
