At what point is the tangent to f(x)=x2+4x−1 horizontal?

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Marin Francis · Accepted Answer

For the tangent to be horizontal, the slope of the graph at that point has to be zero. We find the slope function of f:limh⇒0[(x+h)2+4(x+h)−1]−(x2+4x−1)h=limx⇒0(x+h)2+4(x+h)−1−x2−4x+1h=limh⇒0(x+h)2+4(x+h)−x2−4xh=limh⇒0x2+2hx+h2+4x+4h−x2−4xh=limh⇒02hx+h2+4hh=limh⇒0[2x+h+4]=2x+4Then if the slope is zero:2x+4=0⇒x=−2And so the point of intersection is(-2, f(-2))=(-2,-5)Result:(-2,-5)

hercegvm · Accepted Answer

Use definition for derivative.limh⇒0f(a+h)−f(a)hPlug (a+h) into problem.=[(a+h)2+4(a+h)−1]−(a2+4a−1)hSimplify and combine like terms.=[a2+2ah+h2+4a+4h−1]−(a2−4a−1)hfactor out h.=2ah+h2+4hh=h(2a+h+4)hSolve limit by pluggin in 0.limh⇒02a+h+4=2a+(0)+4=2a+4Horizontal tangent has slope of 0, to set a=0 and solve by subtracting 4 and dividing by 2.2a+4=02a=-4a=-2Plug -2 back into original function to find y value.(−2)2+4(−2)−1=−5Result:(-2,-5)

At what point is the tangent to f(x) = x^2 + 4x -1 horizontal?

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