At first, I didn’t know when to use cosine law and which formula was for solving a side or an angle. My light-bulb moment was realizing that the cosine law is used when you know two sides and the angle between them, or when you know all three sides. The cosine law formula for finding a side length is c² = a² + b² − 2ab cos C. For a simple example, if a triangle has sides a = 6, b = 7, and the angle between them, C = 60 degrees, you plug the values into the formula. That gives c² = 6² + 7² − 2(6)(7)cos60. which is c² = 165. To finish, you take the square root to get c ≈ 12.8. Once I understood when and how to use the cosine law, it made it a whole lot easier to solve for triangles.

For the other instance of cosine law, when you are trying find an angle, lets use the same side angles of a=6, b=7, and now c=8. and if we want to find the biggest side angle which would be side c, because it is opposite from the biggest side. the formula for cosine law for an angle is cosA=b² + c² – a²/2bc. so plugging in the values, we will have cosC= 7² + 6² – 8²/2(7)(6). after plugging those into our calculator, we get cosC= 0.25. and now we got to isolate C, by moving the cos over which gives us an inverse cosine, which we will have C = cos⁻¹(0.25). which gives us around 75 degrees.

This week I realized I didn’t really know how to find the non-permissible values for a rational expression, the right way is to look only at the denominator, set it equal to zero, and solve. For example, with (x+2)(x+2)/(x+5)(x+3), the denominator is (x+5)(x+3), so set (x+5)(x+3) = 0 which gives x = -5 and x = −3. Those values are non-permissible because they make the original denominator zero. So always find zeros of the original denominator first, list them as non-permissible.

For instances where the x in the denominator has a coeficient, for example (2x-3), then you have to put it as an inequality, and try to solve for x by isolating it

as you can see by the end after isolating x, we can see that x connot be 3/2

This week I realized I didn’t actually know how to convert a quadratic function from vertex form into general form. the first step is always to deal with the squared bracket. For example, if the equation is y = 2(x-3)² + 8, the correct way to begin is to rewrite the squared part as (x-3)(x-3). That changes the equation into y = 2(x-3)(x-3) + 8, which puts it in a more expanded form and makes it much easier to find the stretch of the function and also the Y intercept. From there, you just foil the two “(x-3)” and then after distribute the 2 into the quadratic to get rid of the brackets and then combine the +8.

This week, the question I got wrong but learned the most from was about graphing quadratic functions. At first, I didn’t really know how to graph them but the moment came when I learned how to graph a quadratic, all you need is the vertex, the axis of symmetry, the direction the parabola opens. For example, with the function y = x², the vertex is at (0, 0) because nothing is added or subtracted. Since the coefficient of x² is positive, the parabola opens upward. Then you plot simple points on each side, like x = 1 is y = 1, x = −1 is y = 1, x = 2 is y = 4, and x = −2 is y = 4. Connecting these points forms a symmetrical, U-shaped parabola.