# Root functions: asymptotes and equations · Grade 11 · 5 units

> Explanation, worked examples and free practice following the Israeli curriculum.

In Root functions: asymptotes and equations (Grade 11 · 5 units) you practise 6 skills: The sign and symmetry of a function with a root · Different horizontal asymptotes at the two ends · Turning points at the end of the domain · Equations with a root · A vertical tangent · The shortest route.

With a root function, always start with the domain. A rational function leaves out single points, but under a root the expression must not be negative, so the domain is usually an interval or a ray, and sometimes two rays or more. A turning point can also be at an end of the domain, so compare the function’s values at the ends with its values at the points where the derivative is 0.

The root of a square is the absolute value: the root of x squared equals x only when x is not negative. So a function with a root can have different horizontal asymptotes at its two ends: when x is very large, the root of x squared plus a number is close to x, and when x is negative and large in size, it is close to minus x.

The derivative of the root of x is one over twice the root of x. For the root of an expression, also multiply by the derivative of the expression under the root. Where the expression under the root is 0, the formula divides by 0, so look at that point on its own: usually the derivative is not defined there and the graph has a vertical tangent, but not always.

To solve an equation with a root, square both sides. Every solution of the original equation also solves the new one, but not the other way round, so put each solution back into the original equation and check, or show that both sides have the same sign.

**Tip:** Before squaring, write down when the side without the root is not negative: a solution where it is negative is rejected.

## What you learn

### The sign and symmetry of a function with a root

- Example: Is the function f(x) = x√(25 − x²) even, odd or neither?
- Options: Even · Neither even nor odd · Odd
- Answer: Odd
- Solution: The domain is −5 ≤ x ≤ 5 , symmetric about 0. Replace x with minus x: the root does not change and the factor x changes sign, so f(−x) = −f(x) and the function is odd.

### Different horizontal asymptotes at the two ends

- Example: Here is the function f(x) = −3x ÷ √(x² + 9) . What is its horizontal asymptote as x → ∞ ?
- Answer: −3
- Solution: The function is defined for every x, since x² + 9 is always positive. When x is very large, the root is close to x, so the quotient is close to −3 . The asymptote is y = −3 .

### Turning points at the end of the domain

- Example: Here is the function f(x) = x√(3 − x) on the domain 0 ≤ x ≤ 3 . At which value of x does the function take its largest value?
- Answer: 2
- Solution: The domain is 0 ≤ x ≤ 3 . By the product rule the derivative is (6 − 3x) ÷ (2√(3 − x)) , which is 0 at x = 2 . There f(2) = 2 × 1 = 2 , and at both ends f(0) = f(3) = 0 . So the largest value is at x = 2 .

### Equations with a root

- Example: Solve the equation √(x − 7) = 6 .
- Answer: 43
- Solution: The domain is x ≥ 7 . Square both sides: x − 7 = 36 , so x = 43 . Check: √36 = 6 , true.

### A vertical tangent

- Example: The graph of f(x) = √(x + 4) has a vertical tangent at one point. At which value of x ?
- Answer: −4
- Solution: The domain is x ≥ −4 . The derivative is f′(x) = 1 ÷ (2√(x + 4)) . It is not defined at x = −4 , and grows without bound close to it. So at x = −4 , where the graph starts, the tangent is vertical.

### The shortest route

- Example: Two camps, A and B, are on the same side of a straight road: A is 3 km from the road and B is 1 km from it. The points of the road nearest them are 2 km apart. A walker goes in a straight line from A to a point on the road, x km from the point nearest A, and from there in a straight line to B. (A straight road and two camps on the same side of it, A farther from the road than B, and a route from A down to a point on the road and up to B) Which expression is the length of the route, in km?
- Options: √(9 + x²) + √(1 + x²) · √(9 + x²) + √(1 + (2 + x)²) · √(9 + x²) + 2 − x · √(9 + x²) + √(1 + (2 − x)²)
- Answer: √(9 + x²) + √(1 + (2 − x)²)
- Solution: The first leg is the hypotenuse of a triangle with legs 3 and x , so it is √(9 + x²) long. The second is the hypotenuse of a triangle with legs 1 and 2 − x . So the length is √(9 + x²) + √(1 + (2 − x)²) .

The exercises are made by rules and the answers are worked out in code. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 5 units (grades 10 to 12).

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