Radical Functions

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Pre-Calculus › Radical Functions

Questions 1 - 10

Which of the following is and accurate graph of $f(x) = 1 + \sqrt{x + 3}$ ?

Varsity2

Varsity3

Varsity4

Varsity5

Varsity6

Explanation

Remember , for .

Step 1, realize where starts: A) observe never occurs, B) zero-out the radical component of ;

$f(-3) = 1 + \sqrt{-3 + 3}$

$= 1 + \sqrt{0}$

C) The resulting point is .

Step 2, find simple points for after :

, so use ;

$f(-2) = 1 + \sqrt{-2 + 3}$

$= 1 + \sqrt{1}$

The next resulting point; .

, so use ;

$f(1) = 1 + \sqrt{1 + 3}$

$= 1 + \sqrt{4}$

The next resulting point; .

Step 3, draw a curve through the considered points.

Which of the following is and accurate graph of $f(x) = 1 + \sqrt{x + 3}$ ?

Varsity2

Varsity3

Varsity4

Varsity5

Varsity6

Explanation

Remember , for .

Step 1, realize where starts: A) observe never occurs, B) zero-out the radical component of ;

$f(-3) = 1 + \sqrt{-3 + 3}$

$= 1 + \sqrt{0}$

C) The resulting point is .

Step 2, find simple points for after :

, so use ;

$f(-2) = 1 + \sqrt{-2 + 3}$

$= 1 + \sqrt{1}$

The next resulting point; .

, so use ;

$f(1) = 1 + \sqrt{1 + 3}$

$= 1 + \sqrt{4}$

The next resulting point; .

Step 3, draw a curve through the considered points.

Solve the following radical equation.

$\sqrt{x + 2} = 10$

Explanation

When dealing with a radical equation, do the inverse operation to isolate the variable. In this case, the inverse operation of a square root is to square the expression. Thus we square both sides to continue. This yields the following.

Solve the rational equation: $\frac{\sqrt{x}}{\sqrt{2}}=\sqrt{4x+13}$

$x=\frac{-26}{7}$

$x=\frac{26}{7}$

$x=\frac{-3}{13}$

$x=\frac{-13}{3}$

$x=\frac{13}{3}$

Explanation

$\frac{\sqrt{x}}{\sqrt{2}}=\sqrt{4x+13}$

Square both sides to eliminate all radicals:

$\frac{x}{2}=4x+13$

Multiply both sides by 2:

Combine and isolate x:

$x=\frac{-26}{7}$

Solve the rational equation: $\frac{\sqrt{x}}{\sqrt{2}}=\sqrt{4x+13}$

$x=\frac{-26}{7}$

$x=\frac{26}{7}$

$x=\frac{-3}{13}$

$x=\frac{-13}{3}$

$x=\frac{13}{3}$

Explanation

$\frac{\sqrt{x}}{\sqrt{2}}=\sqrt{4x+13}$

Square both sides to eliminate all radicals:

$\frac{x}{2}=4x+13$

Multiply both sides by 2:

Combine and isolate x:

$x=\frac{-26}{7}$

Solve the following radical equation.

$\sqrt{x + 2} = 10$

Explanation

Solve for

$\sqrt{6x+1}=7$

$\frac{13}{6}$

$\frac{64}{6}$

$\frac{50}{6}$

Explanation

In order to get rid of the radical, we square both sides:

$(\sqrt{6x+1})^{2}=7^{2}$

Since the radical cancels out, we're left with

Subtracting both sides by 1 gives us

We then divide both sides by 6 to get .

Solve for

$\sqrt{6x+1}=7$

$\frac{13}{6}$

$\frac{64}{6}$

$\frac{50}{6}$

Explanation

In order to get rid of the radical, we square both sides:

$(\sqrt{6x+1})^{2}=7^{2}$

Since the radical cancels out, we're left with

Subtracting both sides by 1 gives us

We then divide both sides by 6 to get .

Solve for and use the solution to show where the radical functions intersect:

$\sqrt{x + 1} = \sqrt{2x + 2}$

$x = \frac{1}{2}$

Explanation

To solve, first square both sides of the equation to reverse the square-rooting of the binomials, then simplify:

$(\sqrt{x + 1})^{2} = (\sqrt{2x + 2})^{2}$

Now solve for :

The x-coordinate for the intersection point is .

Choose one of the two radical functions that compose the equation, and set the function equal to y. The more simple a function is, the easier it is to use:

$y = \sqrt{x + 1}$

Now substitute into the function.

$y = \sqrt{-1 + 1}$

$y = \sqrt{0}$

The y-coordinate of the intersection point is .

The intersection point of the two radical functions is .

Now graph the two radical functions:

$f(x) = \sqrt{x + 1}$ , $g(x) = \sqrt{2x + 2}$

Varsity1

Solve for and use the solution to show where the radical functions intersect:

$\sqrt{x + 1} = \sqrt{2x + 2}$

$x = \frac{1}{2}$

Explanation

To solve, first square both sides of the equation to reverse the square-rooting of the binomials, then simplify:

$(\sqrt{x + 1})^{2} = (\sqrt{2x + 2})^{2}$

Now solve for :

The x-coordinate for the intersection point is .

Choose one of the two radical functions that compose the equation, and set the function equal to y. The more simple a function is, the easier it is to use:

$y = \sqrt{x + 1}$

Now substitute into the function.

$y = \sqrt{-1 + 1}$

$y = \sqrt{0}$

The y-coordinate of the intersection point is .

The intersection point of the two radical functions is .

Now graph the two radical functions:

$f(x) = \sqrt{x + 1}$ , $g(x) = \sqrt{2x + 2}$

Varsity1

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