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For each of the above the methodology is fairly similar, first try and do it just by looking at it then try the quadratic formula if that doesn't work. At GCSE level I don't think there's any need to worr...
Consider the curve y = 2x^3 - 2x - 12.1) y-intercept. When x=0, y= -12 3) when x tends to infinity...y tends to infinity and when x tends to negative infinity...y tends to negative infinity 4) stationary ...
(2x-1)2 = 4x2- 4x + 1= 4(x2-x)+1The part of the expression which is: 4(x2-x) indicates that the value is a multiple of 4. The number 1 is then added which means...
y^2 = 256/4 = 64therefore, y = sqrt(64) = 8
First complete the square for both x and y. Move all constants to the right hand side. The square root of this is the radius of the circle. The two constants in the completed square bracket show the x and...
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