Tangents and Normals to a Curve

Mathematics · Calculus & Analysis · ages 17-18

Name ______________________ Date ____________

For the curve y = x² + 5 at x = 3, the point of contact is (3, 6).
Circle one: True False
The normal at a point is perpendicular to the tangent at that same point.
Circle one: True False
For the curve y = x³, what is the gradient of the tangent at x = 2?
- 12
- 8
- 6
- 3
For the curve y = x² + 1, what is the y-coordinate of the point where x = 4?
Answer: ______________
Find the equation of the tangent to y = x² + 2x at x = 1.
- y = 4x + 3
- y = 4x - 1
- y = 3x
- y = 4x + 1
Find the equation of the tangent to y = 4/x at x = 2.
- y = x
- y = -x - 4
- y = -x + 4
- y = 2x - 2
Omar is finding the tangent to y = x² at x = 4. He works out dy/dx = 2x, gets 8, and uses the point (4, 8). Where is the slip?
- he should have used dy/dx = x² to find the point
- the point should have been (8, 4) with the coordinates in that order
- the point comes from the curve, so it is (4, 16), and 8 is the gradient
- the gradient at x = 4 is 4, not 8
For the curve y = x³ - 4x, what is the gradient of the tangent at x = 2?
- 4
- 0
- 12
- 8
The tangent to y = x² + 4 at x = 1 crosses the x-axis. At what value of x?
Answer: ______________
Priya works out a tangent gradient of 3 and writes the normal gradient as -3. Where is the slip?
- the normal gradient should have been 3 as well
- the tangent gradient was wrong, so the normal is too
- a normal gradient is always negative, so -3 is fine
- flipping the fraction was skipped, so it should be -1/3

Answer key

For grown-ups. Fold this page away before handing over the rest.

False · The height comes from the curve: 9 + 5 is 14, so the point is (3, 14). The 6 is dy/dx at x = 3, which is the gradient, not the height.
True · That is what a normal is. Both lines pass through the same point on the curve, and they meet there at a right angle.
12 · dy/dx = 3x², and at x = 2 that is 3 times 4, which is 12. The 8 would be the height of the curve there.
17 · The point comes from the curve, so substitute into y: 4² + 1 is 16 + 1, which is 17.
y = 4x - 1 · The point is (1, 3) since 1 + 2 = 3. The gradient is 2x + 2 = 4 at x = 1. Then y - 3 = 4(x - 1) gives y = 4x - 1.
y = -x + 4 · The point is (2, 2) since 4 divided by 2 is 2. Rewriting as 4x⁻¹ gives dy/dx = -4x⁻², which is -1 at x = 2. Then y - 2 = -1(x - 2) gives y = -x + 4.
the point comes from the curve, so it is (4, 16), and 8 is the gradient · His gradient of 8 is right, but he has used it as a height. Putting x = 4 into the curve gives 16, so the point is (4, 16).
8 · dy/dx = 3x² - 4. At x = 2 that is 12 - 4, which is 8.
-1.5 · The point is (1, 5) and the gradient is 2, so the tangent is y = 2x + 3. Setting y = 0 gives 2x = -3, so x = -1.5.
flipping the fraction was skipped, so it should be -1/3 · She changed the sign but never flipped the fraction. The test is a product of -1, and 3 times -3 is -9, not -1.