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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)
How would you use intercepts to find the vertex in a quadratic equation with two x intercepts?
The vertex must be half way between the two x intercepts
Find the value of x in the equation x-40 equals 60?
x=100
Find an equation of variation where y varies directly as x and y equals 500 when x equals 50?
y = 10x
Find an equation of variation where y varies directly as x and y equals 28 when x equals 7?
4x= y
How do you write y equals x minus 4 x plus 2 in vertex form and find the vertex?
The given equation is y = x - 4x + 2 which can be written as y = -3x + 2 This is an equation of a straight line. Therefore it has no vertex and so cannot be written in vertex form.
How do you find the vertex of an equation in standard form?
To find the vertex of a quadratic equation in standard form, (y = ax^2 + bx + c), you can use the vertex formula. The x-coordinate of the vertex is given by (x = -\frac{b}{2a}). Once you have the x-coordinate, substitute it back into the equation to find the corresponding y-coordinate. The vertex is then the point ((-\frac{b}{2a}, f(-\frac{b}{2a}))).
How do you find the vertex of a parabola when the equation is in standard form?
To find the vertex of a parabola in standard form, which is given by the equation ( y = ax^2 + bx + c ), you can use the formula for the x-coordinate of the vertex: ( x = -\frac{b}{2a} ). Once you have the x-coordinate, substitute it back into the original equation to find the corresponding y-coordinate. The vertex will then be at the point ( (x, y) ).
What is the vertex of the equation Y equals 4X squared minus 8Xplus 9?
To find the vertex of the quadratic equation ( Y = 4X^2 - 8X + 9 ), we can use the vertex formula ( X = -\frac{b}{2a} ). Here, ( a = 4 ) and ( b = -8 ), so ( X = -\frac{-8}{2 \times 4} = 1 ). Substituting ( X = 1 ) back into the equation gives ( Y = 4(1)^2 - 8(1) + 9 = 5 ). Therefore, the vertex of the equation is at the point ( (1, 5) ).
How do you find the vertex of an equation on a graph?
To find the vertex of a quadratic equation in the form (y = ax^2 + bx + c), you can use the formula (x = -\frac{b}{2a}) to determine the x-coordinate of the vertex. Once you have the x-coordinate, substitute it back into the equation to find the corresponding y-coordinate. The vertex is then the point ((x, y)) on the graph. For graphs of other types of functions, the vertex may need to be identified through other methods, such as completing the square or analyzing the graph's shape.
How do you find the vertex of the parabola y equals -4x2 - 16x - 11?
You would convert it to vertex form by completing the square. You can also find the optimum value as optimum value and vertex are the same.
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)
How do you find the vertex of an equation in vertex form?
look for the interceptions add these and divide it by 2 (that's the x vertex) for the yvertex you just have to fill in the x(vertex) however you can also use the formula -(b/2a)
Find the vertex of the parabola y equals -2x2 plus 12x - 13?
The points at which the parabola intersects the x axis are 3-sqrt(10)/2 and 3+sqrt(10)/2. The X position of the vertex is in the middle, at 3. The y position, from there, is simply found by substituting 2 for x in the equation. As a result, the vertex is at (3, 5).
How would you use intercepts to find the vertex in a quadratic equation with two x intercepts?
The vertex must be half way between the two x intercepts
What is the vertex of the function y equals x2-4x 1?
To find the vertex of the quadratic function ( y = x^2 - 4x + 1 ), we can use the vertex formula ( x = -\frac{b}{2a} ), where ( a = 1 ) and ( b = -4 ). This gives ( x = -\frac{-4}{2 \cdot 1} = 2 ). Substituting ( x = 2 ) back into the equation, we find ( y = 2^2 - 4(2) + 1 = -1 ). Thus, the vertex of the function is at the point ( (2, -1) ).
The vertex of the parabola below is at the point (-4-2) which equation below could be one for parabola?
-2
