Tangents and normals practice questions.

Determine the equation of the tangent to the curve y = x² − 4x + 1 at the point where x = 3.

• y = 2x + 1
• y = −(1/2)x − 1/2
• y = 2x − 8
• y = 2x − 2

Answer: y = 2x − 8

At x = 3: y = −2 and dy/dx = 2x − 4 = 2. Then y + 2 = 2(x − 3) gives y = 2x − 8. Option with gradient −1/2 is the normal, not the tangent.

The curve y = x³ − x is given. Determine the gradient of the normal at the point where x = 1.

• −1/2
• 2
• −2
• 1/2

Answer: −1/2

dy/dx = 3x² − 1, so the tangent gradient at x = 1 is 2. The normal gradient is the negative reciprocal: −1/2.

Determine the equation of the normal to the curve y = x² at the point (2, 4).

• y = 4x − 4
• y = −4x + 12
• y = −(1/4)x + 4
• y = −(1/4)x + 9/2

Answer: y = −(1/4)x + 9/2

The tangent gradient is 2x = 4, so the normal gradient is −1/4. Then y − 4 = −(1/4)(x − 2) gives y = −(1/4)x + 9/2. Using −4 instead of −1/4 is the common negative-reciprocal error.

Determine the x-coordinate of the point on the curve y = x² − 6x + 2 where the tangent is parallel to the line y = 2x + 5.

• x = 3
• x = 4
• x = 2
• x = 1

Answer: x = 4

Parallel lines have equal gradients, so solve dy/dx = 2: 2x − 6 = 2 gives x = 4. x = 3 is where the tangent is horizontal, not parallel to the given line.

Determine the equation of the tangent to the curve y = 2x³ at the point where x = 1.

• y = 6x − 4
• y = 6x + 2
• y = 6x − 6
• y = −(1/6)x + 13/6

Answer: y = 6x − 4

At x = 1: y = 2 and dy/dx = 6x² = 6. Then y − 2 = 6(x − 1) gives y = 6x − 4. Forgetting to substitute the point correctly produces the wrong intercept.

The tangent is drawn to the curve y = x² at the point (1, 1). Determine where this tangent crosses the x-axis.

• x = 2
• x = −1
• x = 1
• x = 1/2

Answer: x = 1/2

The tangent has gradient 2 and equation y = 2x − 1. Setting y = 0 gives x = 1/2.