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Cory P.
High School Math Teacher
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Pre-Calculus
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Question:

Suppose that I have a quadratic function x^2 where the vertex of the graph is located at (-3,4). Describe the transformations that took place if the parent function x^2 has a vertex located at the origin.

Cory P.
Answer:

1.) Since the vertex of the graph is at (-3,4) we can tell that the function took on two types of transformations. a. Given that the x value of the vertex is negative we see that a horizontal translation left three units have occurred. b. Given that the y value of the vertex is positive we see that a vertical shift/translation up 4 units have occurred. 2.) Since there are no coefficients in front of x^2 or being added or subtracted to the function we only have two transformations.

Trigonometry
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Question:

Suppose that a 25 foot ladder is leaning up against a wall that is perpendicular to the ground, and the base of the ladder is 15 feet from the wall on level ground. What is the degree of the angle formed where the ladder touches the ground?

Cory P.
Answer:

1.) Draw a picture of the situation that we have. a. In this situation we see that we have a right triangle and the hypotenuse is 25 feet and the base of the triangle is 15 feet. 2.) Now that we have the triangle set up we can use right triangle trigonometry to solve for the angle that is formed with the base of the ladder and the ground. a. With the sides given we see that we are using the adjacent leg and the hypotenuse to solve for this angle. b. This means we need to use the inverse cosine function to solve for the angle. c. So now we have cos^-1(15/25) d. After putting this in the calculator we get 53.1 3.) So this means the angle formed between the ladder and the ground is 53.1 degrees.

Algebra
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Question:

Suppose I have a graph represented by this function (5x^2 -25x-30)/(x+1). What values of x create either a hole or a vertical asymptote?

Cory P.
Answer:

First we need to simplify the function. 1.) Start by simplifying the numerator first. a. We see that each value in the numerator is divisible by 5. b. After dividing out a 5 we get [5(x^2-5x-6)] c. Now we can factor the basic trinomial by finding two number that when multiplied together give me -6, and when I add those same two numbers I get -5. d. The only two numbers that would work in this case would be -6 and 1 e. Now I can re-write the numerator as [5(x-6)(x+1)] 2.) Now look at the numerator and the denominator and see if anything looks identical. a. We now see that we have an (x+1) in the numerator and in the denominator. b. Since they are the same we can eliminate them because anything divided by itself is one. c. Since we were able to cross out (x+1) we can set that binomial equal to zero and find out where our hole in the function lies. d. Since we have (x+1)=0 we subtract 1 from both sides and end up with x=-1 3.) Now that there isn't anything left on bottom there cannot be a vertical asymptote.

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