Week 13 blog post

Posted on December 1, 2018 by saejino2016

On this week, we had a team work activity, we graph and solve absolute value functions and equations at the whiteboard. This activity was very fun. We discussed with our team and we helped each other. It was good at teamwork. Also we learned Graphing Reciprocals of Linear functions and Graphing Reciprocals of Quadratic functions. In these parts, the equation is looked like y= $\frac{1}{x}$ Its parents function is y= x. The reciprocal function’s form is to reflect the parents function. For example, y= x+5 Its reciprocal function is y= $\frac{1}{x+5}$ . In reciprocal functions, there are some important things, the graph of y= $\frac{1}{x-1}$ has no x-intercept because $\frac{1}{x-1} is never 0. Also the x-axis is a Horizontal Asymptote of the graph; that is, a horizontal line that the graph approaches.� � � � � y=$ latex \frac{1}{x-1}$ is undefined when x=1. That is, x=1 is a non-permissible value. The line x=1 is a Vertical Asymptote; that is, a vertical line that the graph approaches but never reaches. This asymptote passes through the x-intercept of y=f(x). In a word, Horizontal asymptote is y=0, Vertical asymptote is x-intercept. When we are graphing the reciprocal function, we have to know invariant points. Invariant points is the points when the y is 1 and -1. In linear reciprocal functions, there is one vertical asymptote. But in quadratic functions, there are maximum of two vertical asymptote. When we graphing Reciprocal functions, first we have to graph parents function and to find the points when y= 1 and -1, and finally to find vertical, horizontal asymptote. In linear reciprocal functions, the invariant points is maximum of two but in quadratic reciprocal functions , the invariant points is maximum of four.

Graphing Linear Reciprocal function and Quadratic function.

Week 12 blog post

Posted on November 26, 2018 by saejino2016

On this week, we finished lesson 5 and we practice lesson 5 for test on Thursday. After test on Thursday, we learned new lesson 8. This lesson is about solving absolute value equations. An absolute value function has the form y = |f(x)|, where f(x) is a function. The x-intercept of the graph y = |f(x)| is a critical point. the graph of y = |f(x)| changes direction at this point. We can solve absolute value equations by graphing using the strategies.

Week 11 blog post

Posted on November 19, 2018 by saejino2016

On this week, we learned how to graph Linear inequality. First, we have to check y-intercept and slope. and next, we have to check inequality sign. If any equation has equal sign, we have to draw a solid line, however the equation doesn’t have equal sign, we have to draw broken line. And if we graph the linear inequality, the graph of the equation divides the coordinate plane into two region, and to put the test number into the equation. and if the equation include the part of to include the test number, we have to paint that part.

Ex) Determine whether each point is a solution of the given inequality

4x-3y ≥ 6 A(4,3) 4(4)-3(3) ≥ 6 -> 16-9 ≥ 6 -> 16 ≥ 15 This point is solution.

7y+2x < -10 B(-8,1) 7(1)+2(-8) < -10 -> 7-16 < -10 -> -9 < -10 This point is not solution

-y+5x ≤ 17 C(2,-7) -(-7)+5(2) ≤ 17 -> 7+10 ≤ 17 -> 17 ≤ 17 This point is solution

-3x+8y+5 > 0 D(-5,-3) -3(-5)+8(-3)+5 > 0 -> 20-24 >0 -> -4 > 0 This point is not solution

Also on this week, we learned how to graph Quadratic inequality. First, we have to find the coordinates of vertex, slope(a), x-intercepts, y-intercept,and direction of opening. And check the inequality sign of the quadratic equation. If the equation has equal sign, we have to draw a solid curve, if the equation doesn’t have equal sign, we have to draw a broken curve. And next, to put the test number into the equation.(Test points can be used to determine which region satisfies the inequality.) And also we have to paint the part of to include the test number.

Ex) To graph Quadratic inequality

Week 10 blog post

Posted on November 12, 2018 by saejino2016

On this week, we reviewed chapter 1, 2, 3, 4 and we learned new chapter, Graphing Inequalities and systems of Equations. At chapter 1, we learned arithmetic sequences, arithmetic series, geometric sequences, and geometric series. At chapter 2 , we learned absolute value and about radicals. At chapter 3, we learned solving quadratic equations. At chapter 4, we learned analyzing quadratic functions. The specific details about chapter 1, 2, 3, 4 are on my previous blog post. Also on this week, we took the midterm test on Wednesday. And we started new chapter, this chapter is about graphing inequalities and systems of equations. In section 5.1, we learned solving quadratic inequalities in one variable. The general form are $ax^2$ + bx+ c < 0, $ax^2$ +bx +c > 0 , $ax^2$ +bx + c ≥ 0, and $ax^2$ +bx +c ≤ 0. The x-intercept of the graph of a quadratic function are called the critical values of the corresponding quadratic inequality. These values can be used to illustrate the solution of the inequality on a number line. For example, the solution of $x^2$ +5x -6 < 0. At first you have to factor. If you factor this example, it will be (x+6)(x-1) < 0. And if you illustrate the number line, you can find the solution. The solution is -6 < x < 1.

Ex) Solve each quadratic inequality.

$4x^2$ – 16x + 12 > 0. At first, you have to fine common factor. The common factor is 4. To divide by 4. Then the equation will be $x^2$ – 4x + 3 > 0 and then to factor. (x-3)(x-1) > 0 and to draw the number line, then the solution is 1 < x < 3.

Ex) A tennis ball is thrown upward at an initial speed of 24m/s. The approximate height of the ball, h metres, after t seconds, is given by the equation h=24t- $4t^2$ . Determine the time period for which the ball is higher than 20m.

(24t- $4t^2$ > 20)* -1/4 if you multiply both sides by minus, you must change the direction of the inequality sign. $latex t^2 -6t +5 < 0 and then to change factored form. (t-5)(t-1) < 0 and draw the number line and find the time period for which the ball is higher than 20m.

this is how to solve my self made question.

Week 7 blog post

Posted on October 21, 2018 by saejino2016

On this week, I learned about Interpreting the Discriminant. The interpreting the discriminant helps to find how many roots in equations.

$b^2$ – 4ac It is called the discriminant of the quadratic equation.

The quadratic equation $ax^2$ + bx +c = 0 has:

$b^2$ – 4ac > 0 It has two real roots. Distinct root/real root

$b^2$ – 4ac = 0 It has one root. Equal root

$b^2$ – 4ac <0 It has not root. No root/ unreal root

Also there are rational root and irrational root. In $b^2$ – 4ac, if this number is perfect square number, it is rational number, but if it is not, it is irrational number.

self make questions) How many roots do this equation have?

$4x^2$ + 2x – 7 = 0 $2^2$ – 4(4)(-7) = 0 4+112 = 0 116 > 0 This equation has two real roots.

explain) At first to find the value of a, b, c. The value of a is 4, b is 2, and c is -7. And each number put in $b^2$ – 4ac = 0 Then 4+112=0 we find the value of this equation. 116 is bigger than zero. So 116 > 0 It has two real roots.

In this chapter, we have to understand how many roots in equation when $b^2$ -4ac is bigger than zero, less than zero, and equal with zero.

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Week 13 blog post

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Week 7 blog post