y2 = 32x
y = ±√32x
the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
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)
The vertex must be half way between the two x intercepts
x=100
You can find a vertex wherever two lines (or line segments) meet.
y = 10x
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.
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}))).
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) ).
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.
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)
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)
The vertex must be half way between the two x intercepts
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).
-2
-2-5
To find the vertex of a parabola given its equation in standard form (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 equation to find the corresponding y-coordinate. Thus, the vertex can be expressed as the point ((-\frac{b}{2a}, f(-\frac{b}{2a}))). For parabolas in vertex form (y = a(x-h)^2 + k), the vertex is simply the point ((h, k)).
y = 2x2 + 3x + 6 Since a > 0 (a = 2, b = 3, c = 6) the graph opens upward. The coordinates of the vertex are (-b/2a, f(-b/2a)) = (- 0.75, 4.875). The equation of the axis of symmetry is x = -0.75.