Month: December 2018

Week 16 – Precalc 11

ximena

This week in Precalculus 11, we reviewed and learned about trigonometry. We learned about quadrants, the initial arm, the terminal arm, the rotation angle, reference angle, and the coterminal angle. The rotation angle is formed between the initial arm (x-axis on the positive side) and the terminal arm. The reference angle is the angle touching the x-axis and the terminal arm.

*Reference angle can never be over 90°.

Ex.

To find the reference angle, subtract the bigger number with the x-axis number. In this case, you have to subtract 180° from 200° but hypothetically, if the rotation angle was 320°, you would have to subtract 320° from 360°.

To find the coterminal angle, you have to subtract the rotation angle from 360°. In this example, we subtract 200° from 360°. The coterminal angle is negative because it is going the opposite way (clockwise).

Week 15 – Precalc 11

ximena

This week in precalculus 11, we learned about rational expressions. We learned how to multiply and divide them, add and subtract them, and how to solve them completely. In this post, I will be focusing on the solving aspect.

Note that the equations could result in a linear or quadratic equations and that there are restrictions.

Ex.

\frac{1}{x-4} + \frac{2}{x+4} = \frac{5}{x^2-16}

The first step is to see if the fractions can factor out.

\frac{1}{x-4} + \frac{2}{x+4} = \frac{5}{(x-4)(x+4)}

The next step is to evaluate if you have to multiply by a common denominator or cross multiply.

In this case, we cannot cross multiply since there are 3 expressions, 2 on one side of the equal sign and one on the other, which means that we have to use the multiplying by a common denominator method. you have to multiply the top and bottom of the fractions by the same number(s)/expression(s) (common denominator) and when you multiply the denominator by the same expression, it cancels out.

Common denominator: (x-4)(x+4)

\frac{1}{x-4} x (x-4)(x+4)

+ \frac{2}{x+4} x (x-4)(x+4)

= \frac{5}{(x-4)(x+4)} x (x-4)(x+4)

 

\frac{1}{x-4} x (x-4)(x+4)

+ \frac{2}{x+4} x (x-4)(x+4)

= \frac{5}{(x-4)(x+4)} x (x-4)(x+4)

Once the expressions are canceld out, multiply the numerator by the remaining expression, if there is one.

(x+4) + 2(x-4) = 5

From here, you expand, add or subtract like terms, then solve.

x + 4 + 2x – 8 = 5

3x -4 = 5

3x = 9

\frac{3x}{3} = \frac{9}{3}

x = 3

State the non-permissable values.

x ≠ 4, -4

Week 14 – Precalc 11

ximena

This week in precalculus 11, we learned about equivalent reciprocal functions.

Recall that a rational number is a quotient of two integers, except the denominator cannot equal 0. The quotient of two polynomials is called a rational expression.

Equivalent rational expressions: \frac{1}{2}, \frac{2}{4}, \frac{15}{30}, etc.

When trying to find equivalent rational expressions you factor it out, then simplify it by canceling out the expression that is in the top and bottom of the fraction.

Since the denominator can never equal 0, you have to find the non-permissible values.

Ex.

\frac{2x+2}{x^2+3x+2}

= \frac{2(x+1)}{(x+2)(x+1)}

x ≠ -2, -1

With this equation, you can cancel out (x+1) from the top and bottom of the equation.

= \frac{2}{x+2}

x ≠ -2

Week 13 – Precalc 11

ximena

This week in Precalculus 11, we learned about Reciprocal Functions.

There are 4 steps to graphing them:

First, graph the linear or quadratic parent function. Once the parent function is graphed, identify the invariant points – which is always 1 and -1 on the y-axis. Next, identify the asymptotes. The horizontal asymptote is always y = 0 and the vertical asymptote is placed at the x-intercept. Lastly, draw the hyperbola, in which you start at the invariant points and draw the lines closer and closer to the asymptotes but never touching or crossing the asymptotes. There are no x-intercepts, as the hyperbolas don’t ever touch the x-axis.