Yes. Think of y as being a function of x. y = f(x) = x2 + 1
If: y = x2-7x+8 and y = -x2+9x-6 Then: x2-7x+8 = -x2+9x-6 So: 2x2-16x+14 = 0 => x2-8x+7 = 0 Therefore: x = 1 and x = 7 By substitution: x =1, y = 2 and x = 7, y = 8 Points of intersection: (1, 2) and (7, 8)
(the shape is an upside down 'u').
y = x2 - 4x + 4 can be factored into y = (x-2)(x-2) The repeated factor is 2.
y = x2 + 8x + 15 = (x + 3)(x + 5). When y = 0, one or other of the two factors must equal zero; that is, x = -3 or -5, when y = 0.
y ≥ 11
Y=(x+8)^2-5
is a quadratic equation for y, in terms of x.
If: y = x2+20x+100 and x2-20x+100 Then: x2+20x+100 = x2-20x+100 So: 40x = 0 => x = 0 When x = 0 then y = 100 Therefore point of intersection: (0, 100)
the graph is moved down 6 units
The vertex is at (-1,0).
Yes. Think of y as being a function of x. y = f(x) = x2 + 1
y = x2 + 4 The graph is a parabola, with its nose at y=4 on the y-axis, and opening upward.
-2-5
No, that's a parabola.
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
one