The expression "y squared plus y" can be written mathematically as (y^2 + y). This expression represents a quadratic equation in terms of (y). If you're looking to set it equal to something, you would typically write it as (y^2 + y = 0) or (y^2 + y = k), where (k) is any constant. To solve for (y), you could factor or use the quadratic formula, depending on the context.
89
2
(x, y) = (-3, -3) or (3, 3)
k = 0.1
They touch each other at (0, 100) on the x and y axis.
89
2
(x, y) = (-3, -3) or (3, 3)
k = 0.1
They touch each other at (0, 100) on the x and y axis.
y = x - 4 x^(2) + y^(2) = 8 Hence y^(2) = 8 - x^(2) Square up the linear eq'n and substitute Hence y^(2) = ( x - 4)^(2) NB Notice the RHS is wholly squared , not individual terms. Substitute ; this eliminates 'y' ( x - 4)^(2) = 8 - x^(2) Square out the brackets x^(2) - 8x + 16 = 8 - x^(2) Collect terms 2x^(2) -8x +8 = 0 Factor 2(x^(2) - 4x + 4) = 0 (x- 2)(x - 2) = 0 Hence x = 2 ( The 'x' coordinate) . Substitute into either eq;n for 'y' y = 2 - 4 y = -2 Hence ( x,y) = ( 2,-2) ; The point of contact of the tangent and the curve.
y equals x plus 4 when y equals 0 then x equals 2i i is the square root of negative 1
They are +/- 5*sqrt(2)
Two equations: a linear one in y and a quadratic in x.
y=xsquared-4x+2
They intersect at the point of: (-3/2, 11/4)
It is the equation of a parabola.