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Find the tangent to the curve y=x^2 +2x at point (1,3)

In order to find the gradient we need to differentiate d/Dx = 2x + 2using our point, the gradient is 4using y = mx+cy = 4x +cusing our pointsy= 4x - 1
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Answered by Diana W. • Maths tutor
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What are the limits of an inverse tan graph.

pi/2 and -pi/2
EH
Answered by Emily H. • Maths tutor
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Solve x^3+2*x^2-5*x-6=0

First find a root of the function: f(x)=x^3+2x^2-5x-6 f(2)=(2)^3+2*(2)^2-5*(2)-6 =0. Therefore, (x-2)(x^2+Ax+3)=0 where A is an unkown constant. Compare x^2 coeffecients: A-2=2, A=4. So (x-2)(x^2+4x+3)=0. Th...
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A new sports car accelerates using rockets at 5m/s for 30 seconds from some traffic lights and then decelerate for 45 seconds to a stop.

SUVAT Part (A) Part (B) S=? S=?U=0 U= Calc From A (V)V=? V= 0T=30s t=45sPart A: V=U+atV=0+5 30V= 150S=ut+at 2 S=0+5 30*30S=4500mPart B: S=(U+V) t/2S=75 45S=3375mTotal Distance = 3375+4500=7875
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Answered by Matthew W. • Maths tutor
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Express cos(2x) in terms of acos^2(x) + b

cos(2x)=cos 2 (x) - sin 2 (x)use sin 2 (x) = 1 - cos 2 (x)so cos(2x)=cos 2 (x) - (1 - cos 2 (x))cos(2x)=2cos 2 (x) - 1so a = 2 and b = -1
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