Find the tangent to the equation y=x^2 -2x +4 when x=2

When X=2 Y=2^2-2(2)+4=4 So the coordinates are (2,4)Differentiate Y so dy/dx = 2x-2Tangent Gradient when x=2 is 2(2)-2=2 so m=2We need to find the y intercept to get out tangent equationso y=2x+c , we sub in our coordinates to get 4=2(2)+c , c=0So therefore the tangent equation is y=2x

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Answered by Nabeel N. Further Mathematics tutor

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This is a question from a past paper: https://prnt.sc/r6jnxc


Prove that tan^2(x)=1/(cos^2(x))-1


A curve is mapped by the equation y = 3x^3 + ax^2 + bx, where a is a constant. The value of dy/dx at x = 2 is double that of dy/dx at x = 1. A turning point occurs when x = -1. Find the values of a and b.


How would you differentiate x^x?


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