During week 15 we learnt how to add and subtract radical expressions, in the picture above I demonstrate the part of addition
For the first example, There is an expression with the denominators being the same so all you have to do is add across. An important note, to subtract or add your denominator must be the same for both sides.
For the second expression the denominators are different so what was done in the example to find the lowest common multiple is we’ve multiplied each side with the opposite denominator so that the denominators become the same and then you simply add across
equivalent rational expressions are basically factoring in a fraction format and then once you have the expressions factored as I have in the second step of the expression and after that you must find the restriction where the denominator cannot be equal to 0 and the restrictions are -5 and 0
after that you cancel the like terms from the nominator and denominator which in this case is (x+5) and the final expression will be the one on the bottom
this week during math we learnt how to graph reciprocal linear and quadratic functions, in the picture above represented are 3 different ones. The first being linear the second being a simple quadratic function and the final being a more complex quadratic function.
the key to reciprocal functions is making sure you know exactly where your asymptotes are located and they will correspond with your interveniant numbers which are 1 and -1 and with your restrictions they represent where the hyperbolas will be located.
so for the first one x=3 and y=0 those will be our restrictions and the hyperbolas are set.
for the second one it’s more difficult because it’s a quadratic function but basically your asymptotes are what give you boundaries or sections and usually with a quadratic function it will be 6 sections as to which the parabola could be located in and simply which ever section the parabola is found in that’s where and which direction you draw the hyperbola the same goes for the bottom quadratic function
This week in math we learnt how to determine x-intercepts using the substitution method.
the very first step is you have to isolate a variable, in the equations above a variable has already been isolated for us which is “y” and from that step is the second where you substitute the letter variable which is “y” in the second equation to the equation above.
after that step 3 is distributing and and putting the equation in a form you can factor and then factor it to the point of being able to determine the x-intercepts and after that you do the math and the x-intercepts should be -5/4 and 1 for these equations
this week we learnt how to solve quadratic inequalities and how to test if the equation represented is true
above we have a standard quadratic equation but instead of being = to it is saying that the equation is greater than 0
so to test that we first have to find the x intercepts and from there we find a number lower than the lowest x intercept, a number in between the x intercepts, and a number greater than the greatest x intercept and we input them in the equation and see if the answer given is a positive or negative which will then tell us if the answer is greater or less than 0
we have now tested all three of the numbers we chose and it says that we have a positive, a negative, and another positive answer. We are looking for the positive answers to make this equation true.
after that you just state what is true which is….. X<-4 and x>2 making the equation true