02Esqrt

toc = Square Root Function =

Power Functions
A power function has the form y = x n.
 * For n = 1, we get a linear graph,
 * For n = 2, we get a parabola,
 * For n = 3, we get a cubic,
 * For n = 4, we get a quartic,
 * For n = –1, we get a hyperbola,
 * For n = –2, we get a truncus,
 * For n = ½, we get the graph of a ** square root function **.

math y = x^{\frac{1}{2}} \; \text{ is more familiar as } y = \sqrt{x} \qquad. math

... ... Domain: x Î R+ È { 0 } ... ... (or) ... ... Domain: { x: x ≥ 0 }

... ... Range: y Î R+ È { 0 } ... ... (or) ... ... Range: { y: y ≥ 0 }

... ... Intercepts: ( 0, 0 ) (at the origin)

Notice that if we combine the positive and negative versions of the square root function, we get the graph of a relation that is a parabola drawn sideways.

math . \qquad \text{This is because } y = \pm \sqrt{x} math ... ..... is equivalent to x = y 2 , ... ..... which is the **inverse** of y = x 2.

... ... OR
 * This means that they are a reflection across the line y = x
 * the x and y coordinates of individual points are swapped
 * so (x, y) → (y, x)

We can apply the standard transformations such as dilations, translations and reflections when sketching a square root function.

** Example 1 **
math . \qquad \text{Sketch } y = \sqrt{-2x-2}-1 \qquad. math


 * Solution:**

... ... This can be written as:

math . \qquad y=\sqrt{-2 \left(x+1 \right) }-1 \qquad. math

This is the standard square root function with the following transformations.
 * Reflected across the y-axis (in the x-direction)
 * Dilated by a factor of ½ in the x-direction
 * Translated to the left by 1 unit
 * Translated down by 1 unit

Domain:

math . \qquad \big\{ x: \; x \leqslant -1 \big\} \qquad. \\ . \\ . \qquad \text{or } \\. \\ . \qquad x \in \big( -\infty, \; -1 \big] math

Range: math . \qquad \big\{ y: \; x \geqslant -1 \big\} \qquad. \\ . \\ . \qquad \text{or } \\. \\ . \qquad y \in \big[ -1, \; \infty \big) math

... ... There are no stationary points.

... ... There is an x-intercept at x = –1.5

... ... There are no y-intercepts

Note:
 * a negative sign outside the sqrt reflects the graph in the y-direction (across the x-axis)
 * a negative sign with the x inside the sqrt reflects the graph in the x-direction (across the y-axis)

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 * the domain can always be found by setting the insides of the sqrt to __>__ 0 then solving for x.

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