Math - Algebra II | Practice Hub

$y=\frac{3x^2-5}{2x^2+7}$

What are the horizontal asymptotes of this equation?

$y=\frac{3}{2}$

There are no horizontal asymptotes.

$y=\frac{2}{3}$

Explanation

Since the exponents of the variables in both the numerator and denominator are equal, the horizontal asymptote will be the coefficient of the numerator's variable divided by the coefficient of the denominator's variable.

For this problem, since we have $\frac{3x^2}{2x^2}$ , our asymptote will be $y=\frac{3}{2}$ .

$y=\frac{x^2-9}{x^2-16}$

What is the horizontal asymptote of this equation?

There is no horizontal asymptote.

Explanation

Look at the exponents of the variables. Both our numerator and denominator are . Therefore the horizontal asymptote is calculated by dividing the coefficient of the numerator by the coefficient of the denomenator.

$y=\frac{1}{1}=1$

Solve for :

$\left | 2x-5\right |=3x$

Explanation

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

$\left | 2x-5\right |=3x$

and

This gives us:

and

$x=-5\ and\ 1$

However, this question has an outside of the absolute value expression, in this case . Thus, any negative value of will make the right side of the equation equal to a negative number, which cannot be true for an absolute value expression. Thus, is an extraneous solution, as $\left | 2x-5\right |$ cannot equal a negative number.

Our final solution is then

Simplify $3\frac{3}{8}\cdot 2\frac{1}{3}$ .

$7\frac{7}{8}$

$6\frac{1}{8}$

$5\frac{5}{8}$

$6\frac{2}{3}$

$8\frac{1}{3}$

Explanation

Chenge the mixed numbers into improper fractions by multiplying the whole number by the denominator and adding the numerator to get

$\frac{27}{8}\cdot \frac{7}{3}=\frac{189}{24}=\frac{63}{8}; or; 7\frac{7}{8}$ .

Solve for :

$\sqrt{x^2+12x+36} =4$

$x = \left { -10, -2\right }$

Explanation

To solve for in the equation $\sqrt{x^2+12x+36} =4$

Square both sides of the equation

$\rightarrow (\sqrt{x^2+12x+36})^2 =(4)^2$

$\rightarrow x^2+12x+36 =16$

Set the equation equal to by subtracting the constant from both sides of the equation.

$\rightarrow (x^2+12x+36) - 16 =(16) -16$

$\rightarrow x^2+12x+20 =0$

Factor to find the zeros:

$\rightarrow (x+10)(x+2) = 0$

This gives the solutions

$x = \left { -10, -2\right }$ .

Verify that these work in the original equation by substituting them in for . This is especially important to do in equations involving radicals to ensure no imaginary numbers (square roots of negative numbers) are created.

Solve using the quadratic formula:

$a=\frac{5 \pm i\sqrt{7}}{4}$

$a=\frac{6 \pm i\sqrt{7}}{4}$

$a=\frac{4 \pm i\sqrt{7}}{4}$

$a=\frac{3 \pm i\sqrt{7}}{4}$

$a=\frac{7 \pm i\sqrt{7}}{4}$

Explanation

Use the quadratic formula to solve:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$a = \frac{-(-5) \pm \sqrt{(-5)^2-4(2)(4)}}{2(2)}$

$a = \frac{5 \pm \sqrt{-7}}{4}$

$a = \frac{5 \pm i\sqrt{7}}{4}$

Find the center and radius of the circle defined by the equation:

$\left ( 3,-5 \right )\ and\ r=6$

$\left ( 3,5 \right )\ and\ r=6$

$\left ( -3,-5 \right )\ and\ r=18$

$\left ( -3,5 \right )\ and\ r=18$

$\left ( -3,5 \right )\ and\ r=6$

Explanation

The equation of a circle is: $(x-x_{1})^2+(y-y_{1})^2=r^2$ where is the radius and $(x_{1},y_{1})$ is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to 0, which is 3. We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 6.