If (2,6) lies on the curve f(x)=ax2+bx and y=x+4 is a tangent to the curve at that point. Find a and b?

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Buizzae77t · Accepted Answer

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.We have:f(x)=ax2+bxWe are given that the point (2,6) lies on the curve:⇒f(2)=6∴4a+2b=6∴2a+b=3...[A]If we differentiate the parabola equation, then we have:f&#039;(x)=2ax+bThe gradient of the tangent at (2,6) is given by:mT=f′(2)=4a+bWe also know that the tangent equation is y=x+4 and so comparing with the standard straight line equation y=mx+c:⇒mT=1∴4a+b=1...[B]We now solve the equations [A] and [B] simultaneous:[B]-[A]⇒2a=−2⇒a=−1Substitute a=-1 into [A]:b−2=3⇒b=5Hence we have:a=-1, b=5

If (2,6) lies on the curve \(\displaystyle{f{{\left({x}\right)}}}={a}{x}^{{{2}}}+{b}{x}\)

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