Apply tangent and normal geometry to a polynomial curve

For the tangent to f at x = 2, which quantity gives the gradient?

Change the tangent gradient through all four settings and compare the normal gradient

What remains fixed?

Tangent and normal lines

The point on a curve is found from its function value, and the tangent gradient is found from its derivative. A normal line has the negative reciprocal gradient when the tangent gradient is non-zero

1. 1Find the point At x = a, the point on the curve has coordinates (a, f(a)) \(\displaystyle P=\left(a,f(a)\right)\)
2. 2Find the tangent gradient Differentiate the function and substitute the x-coordinate of the point \(\displaystyle m_{\mathrm{t}}=f'(a)\)
3. 3Find the normal gradient A normal is perpendicular to a non-horizontal tangent, so its gradient is the negative reciprocal \(\displaystyle m_{\mathrm{n}}=-\frac{1}{m_{\mathrm{t}}}\)
4. 4Use point-gradient form Substitute the point and the required gradient into the equation of a straight line \(\displaystyle y-y_0=m\left(x-x_0\right)\)

Given

\(\displaystyle f(x)=x^2+2x+1,\quad x=2\)

Calculate the y-coordinate when x = 2.

Given

\(\displaystyle f(x)=x^2+2x+1,\quad x=2\)

Differentiate f and calculate the tangent gradient at x = 2.

Using the function value as the gradient. One of these lines is right. Which one?

Normal through the contact point

For the example at x = 2, the curve height is 9 and the tangent gradient is 6. The normal passes through (2,9) and has gradient −1/6.

1. 1Find the normal gradient A normal is perpendicular to a non-horizontal tangent, so its gradient is the negative reciprocal \(\displaystyle m_{\mathrm{n}}=-\frac{1}{m_{\mathrm{t}}}\)
2. 2Use point-gradient form Substitute the point and the required gradient into the equation of a straight line \(\displaystyle y-y_0=m\left(x-x_0\right)\)

Given

\(\displaystyle m_{\mathrm t}=6\)

Find the normal gradient when the tangent gradient is 6.

At a point where the tangent is horizontal, what is the normal?

Given

\(\displaystyle P=(2,-3)\)

The contact point is (2,−3) and its tangent is horizontal. Enter a for the normal x = a.

Find the contact from a line condition

A tangent parallel to y = 4x − 7 must have gradient 4. For f(x) = x² + 2x, solve 2x + 2 = 4, then use the original function to find the height

1. 1Translate a line relation into a derivative condition Parallel nonvertical lines have equal gradients. Perpendicular nonvertical lines have gradients whose product is −1. Solve the resulting equation for every contact input

Given

\(\displaystyle 2x+2=4\)

Find the input where this tangent is parallel to a line of gradient 4.

For a cubic whose derivative is 3x² − 3, which inputs have horizontal tangents?

A tangent can meet a curve again

For f(x) = x³ − 2x + 2, the tangent at x = 1 is y = x. Subtract the line from the curve: x³ − 3x + 2 = (x − 1)²(x + 2). The repeated root identifies the contact; the other factor identifies a second intersection

1. 1An intersection uses the curve and the line Set the original curve equal to the tangent line, then factor. The tangency input may be a repeated root; count it as one point and retain any other distinct roots

Given

\(\displaystyle (x-1)^2(x+2)=0\)

Find the other intersection input, excluding the contact x = 1.

Canonical line coefficients

The line can be written in general form as shown. The integer coefficients have no common factor and A is positive. The retained coefficient table instead prints Ax + By = C and requires primitive integers with B negative. Follow the form and sign convention printed in each question

1. 1State a unique general form For Ax + By + C = 0, clear denominators, remove the common integer factor, and choose A positive, or B positive when A is zero

Given

\(\displaystyle 2x-3y+C=0\quad\text{for }y=\frac23x-\frac43\)

Enter the constant C in the stated left-hand-side general form

An axis intercept of the normal

For contact (1,3) and tangent gradient 2, use the normal equation shown. At the horizontal axis, y is zero. Substitute zero, then solve for x.

1. 1Find the normal gradient A normal is perpendicular to a non-horizontal tangent, so its gradient is the negative reciprocal \(\displaystyle m_{\mathrm{n}}=-\frac{1}{m_{\mathrm{t}}}\)
2. 2Use point-gradient form Substitute the point and the required gradient into the equation of a straight line \(\displaystyle y-y_0=m\left(x-x_0\right)\)

Given

\(\displaystyle -3=-\frac12(x-1)\)

Find the horizontal-axis intercept input of this normal