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

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have the function y = x^2.

Now the derivative of a curve at a particular point gives the slope of the tangent at that point.

Now, we use the relation that the derivative of x^n is n*x^ (n-1)

Here y = x^2, so y’ = 2x.

Now the tangent at the point (-1/2, 1/4) is a line with slope equal to 2*(-1/2) = -1.

The equation of this line is y – 1/4 = -1(x – (-1/2)

=> y – 1/4 = -x – 1/2

=> y + x + 1/4 =0

=> 4x + 4y +1 =0

The tangent at the point (1, 1) has a slope equal to 2*1 = 2

The equation of this line is y – 1 = 2*(x – 1)

=> y - 1 = 2x -2

=> y -2x +1 = 0

giorgiana1976's profile pic

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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

The tangent to the curve y = f(x) dx at (x1,y1) is given by:

y - y1 = m(x1,y1).(x-x1)

Where m(x,y) = dy/dx = (x^2) = 2x.

Therefore m(x1,y1) = m(-1/2 , - 1/4) = 2(-1/2) = -1.

Therefore, tangent at (-1/2,1/4) to  y= x^2 is:

y-1/4 = -1{x-(-1/2)}

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

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

y+x+1/4 = 0

4x+4y+1 = 0 is the tangent at (-1/2), 1/4).

To find the tangent at (1,1):

m(x1,y1) = 2(1) = 2.

Therefore tagent at (1,1) is:  y -1 = 1(x-1)

y-1 = x-1

y = x.

Or x= y, or x-y = 0 is the tanget at (1,1) to y = x^2.

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