Find the gradient of the tangent to the line y=(x-2)^2 at the point that it intercepts the y-axis
First find the coordinates of the point in question:We know x=0By plugging this into the equation of the line we get y=(0-2)2 = (-2)2 = 4Therefore the point is (0,4)
To find the gradient of a line, we differentiate the equation of the line:By substitution -> y=u2 , u=x-2dy/dx=dy/du.du/dxdy/du = 2u , du/dx=1Therefore dy/dx =2u=2x-4Subbing in known coordinate into this equation we get:dy/dx(x=0,y=4) = -4Answer = -4
AJ
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