How do you find the equation of the tangent and normal line to the curve y=6-x^2y=6−x2 at (2,2)?

1 Answer
Steve M
Mar 22, 2017

Tangent:

y = -4x+10 y=−4x+10

Normal:

y = 1/4x + 3/2 y=14x+32

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent so the product of their gradients is -1−1

We have:

y = 6-x^2 y=6−x2

First let us check that (2,2)(2,2) lies on the curve:

x=2 => y=6-4=2 x=2⇒y=6−4=2

Then differentiating wrt xx gives us:

dy/dx = -2x dydx=−2x

When x = 2 => dy/dx = -4 x=2⇒dydx=−4

So the tangent passes through (2,2)(2,2) and has gradient m_T=-4mT=−4, and the normal has gradient m_N=1/4mN=14 so using the point/slope form y-y_1=m(x-x_1)y−y1=m(x−x1) the equation we seek are;

Tangent:

y - 2 = -4(x-2) y−2=−4(x−2)
:. y - 2 = -4x+8
:. y = -4x+10

Normal:

y - 2 = 1/4(x-2)
:. y - 2 = 1/4x-1/2
:. y = 1/4x + 3/2

We can confirm this solution is correct graphically:
enter image source here

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