tangentsFind equations for the tangents to the curve y = x^2 at the points (-1/2 , 1/4) and (1,1).

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The equation of the tangent to the curve, in the specified points is:

y - 1/4 = y'(-1/2)(x + 1/2)

Since y = x^2

We'll differentiate both sides with respect to x:

dy/dx = d/dx(x^2)

dy/dx = 2x

For x = -1/2 => y' = 2*(-1/2) = -1

The equation of the line is:

y - 1/4 = -(x + 1/2)

We'll remove the brackets:

y - 1/4 = -x - 1/2

We'll move all terms to the left side:

x + y - 1/4 + 1/2 = 0

x + y + 1/4 = 0

The equation of the tangent line, at the point  (-1/2 , 1/4), in the general form, is:

4x + 4y + 1 = 0

The equation of the tangent line, at the point (1,1), is:

y - 1 = y'(1)(x-1)

For x = 1 => y'(1) = 2*1 = 2

y - 1 = 2(x-1)

We'll remove the brackets:

y - 1 = 2x - 2

We'll move all terms to the left side:

-2x + y + 1 = 0

2x - y - 1 = 0

The equation of the tangent line, at the point (1,1),in the general form, is:

2x - y - 1 = 0

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