8
The vertex is at the point (0, 4).
A common technique to rewrite a quadratic function in standard form ( ax^2 + bx + c ) to vertex form ( a(x - h)^2 + k ) is called "completing the square." This involves taking the coefficient of the ( x ) term, dividing it by 2, squaring it, and then adding and subtracting this value inside the function. By rearranging, you can express the quadratic as a perfect square trinomial plus a constant, which directly gives you the vertex coordinates ( (h, k) ).
(-3, -5)
The only thing you can do is to rewrite it as x(x + 20). For anything else, you need the value of x.
(x + 5)2
The vertex is at the point (0, 4).
A common technique to rewrite a quadratic function in standard form ( ax^2 + bx + c ) to vertex form ( a(x - h)^2 + k ) is called "completing the square." This involves taking the coefficient of the ( x ) term, dividing it by 2, squaring it, and then adding and subtracting this value inside the function. By rearranging, you can express the quadratic as a perfect square trinomial plus a constant, which directly gives you the vertex coordinates ( (h, k) ).
(-3, -5)
The only thing you can do is to rewrite it as x(x + 20). For anything else, you need the value of x.
Y = X2 + 6X + 2set to 0X2 + 6X + 2 = 0X2 + 6X = - 2now, halve the linear coefficient ( 6 ), square it and add it to both sidesX2 + 6X + 9 = - 2 + 9gather terms on the right and factor the left(X + 3)2 = 7(X + 3)2 - 7 = 0==============vertex form(- 3, - 7)=======vertex
(x + 5)2
The vertex of the positive parabola turns at point (-2, -11)
The standard form of the quadratic function in (x - b)2 + c, has a vertex of (b, c). Thus, b is the units shifted to the right of the y-axis, and c is the units shifted above the x-axis.
104
The equation you gave, 2y+5 = (x-4)2 can also be expressed as y = 0.5(x-4)2 - 5. In the form a(x-p)2 + q, the vertex is the point (p, q). Thus, the vertex of 2y + 5 = (x-4)2 is (4, -5).
23
No