Scottish Higher Maths / Notes

Tangents

Use a derivative and a point on a curve to determine the equation of its tangent.

On this page

What a tangent tells you

A tangent is a straight line that has the same gradient as a curve at one chosen point. The derivative supplies that gradient; the original curve supplies the point the line must pass through.

The tangent method

Use the same four steps each time: differentiate the curve; evaluate the derivative at the stated xx-value; find the matching yy-coordinate from the curve; then substitute the gradient and point into point-gradient form and simplify if requested.

Worked example

A polynomial curve

Find the equation of the tangent to y=x2+3x−4y=x^2+3x-4 where x=1x=1.

  1. Step 1

    Differentiate

    rac{dy}{dx}=2x+3.

  2. Step 2

    Find the gradient

    At x=1x=1, m=2(1)+3=5m=2(1)+3=5.

  3. Step 3

    Find the point

    At x=1x=1, y=1+3−4=0y=1+3-4=0, so the point is (1,0)(1,0).

  4. Step 4

    Form the line

    y−0=5(x−1)y-0=5(x-1), so y=5x−5y=5x-5.

Final answer

y=5x−5y=5x-5

The derivative value is the gradient, while the original function value locates the point.

Common mistake

Do not use the curve's yy-coordinate as the gradient.

Equivalent line equations

The same tangent can be written in point-gradient, slope-intercept or general form. For example, y−2=3(x−1)y-2=3(x-1), y=3x−1y=3x-1 and 3x−y−1=03x-y-1=0 all describe the same line. Rearrangement must preserve every sign and constant.

Worked example

When the point is already given

The point (2,11)(2,11) lies on y=3x2−4x+7y=3x^2-4x+7. Find the tangent there.

  1. Step 1

    Differentiate

    rac{dy}{dx}=6x-4.

  2. Step 2

    Evaluate

    At x=2x=2, m=8m=8.

  3. Step 3

    Use the given point

    y−11=8(x−2)y-11=8(x-2).

Final answer

y=8x−5y=8x-5

There is no need to recalculate the yy-coordinate when the contact point is supplied.

Curves using an earlier differentiation method

Some Tangents questions use a curve whose derivative needs an already-learned method such as the chain rule. That method changes how you obtain rac{dy}{dx}, but the tangent construction remains the same.

Worked example

A composite curve

Find the tangent to y=(2x+1)2y=(2x+1)^2 where x=1x=1.

  1. Step 1

    Differentiate

    By the chain rule, rac{dy}{dx}=4(2x+1).

  2. Step 2

    Find the gradient and point

    At x=1x=1, m=12m=12 and y=9y=9, so the point is (1,9)(1,9).

  3. Step 3

    Form the tangent

    y−9=12(x−1)y-9=12(x-1).

Final answer

y=12x−3y=12x-3

The chain rule is used only to obtain the gradient function; the assessed Tangents method follows afterwards.

Self-checkQuick self-check

Find the equation of the tangent to y=x3−2x2+4y=x^3-2x^2+4 at x=2x=2.

Orthic — Learn with Precision