Category Archives: Math 11

Week 17 blog post

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On this week, we reviewed trigonometry because after winter break, almost classmates may forget the content of trigonometry. This chapter is hardest for me. Basically, we should know new vocabularies to solve the problem. These are standard position, terminal arm , terminal point, reference angle, Sine Law, and Cosine Law. Also we should know the formula ; Sin θ = \frac{opposite}{hypotenuse}, Cos θ = \frac{adjacent}{hypotenuse}, Tan θ = \frac{opposite}{adjacent}. We can find to be not given side and angle with that formula. When the angle θ, between 0° and 360°, is measured counterclockwise from the positive x-axis, the angle is in standard position. Also we should know about all quadrants. Quadrant 1 : +sin, +cos, +tan. Quadrant 2 : +sin, -cos, -tan. Quadrant 3 : -sin, -cos, +tan. Quadrant 4 : -sin, +cos, -tan. The plus sign and minus sign is important in this chapter. The reference angle for all 4 angles is the acute angle that the terminal arm makes with the x-axis. For example here are all the angles in standard position have a reference angle of 60°. Then in quadrant 1 has 60° in standard position. Quadrant 2 has 120° in standard position. Quadrant 3 has 240° in standard position. Quadrant 4 has 300° in standard position. Also there is a formula that find reference angle to use standard position and find standard position to use reference angle. In quadrant 1 the reference angle is standard position. In quadrant 2 If you want find the reference angle, 180° –   θ = reference angle, and if you want to find the standard position, 180° – reference angle =  θ. In quadrant 3 If you want to find the reference angle,  θ – 180° = reference angle, and if you want to find the standard position, 180° + reference angle = θ . The last in quadrant 4, if you want to find the reference angle, 360° –  θ = reference angle, and if you want to find the standard position, 360° – reference angle = θ. And we also learned about Sine Law and Cosine Law. In these lessons, there are also formula. In Sine Law, \frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} we use this formula when we want to find missing side. \frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c} we use this formula when we want to find missing angle. In Cosine Law, A^2 = b^2 + c^2 – 2ab Cos A, B^2 = a^2 + c^2 -2ac Cos B, C^2 = a^2 + c^2 -2ab Cos C. We use the Sine Law when we know 2 sides and non included angle, also we use the Sine Law when we know 2 angles and non included side. We use the Cosine Law when we know two sides and the angle between them, and when we know all three sides. If we know basic of this chapter, we can solve all problems.

Thank you 

 

 

 

 

Week 15 blog post

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On this week, we finished chapter 7. In this chapter, lesson 6 is difficult part for me. In lesson 6, we have to understand the question and we can solve the question. Also we have to know about Time= distance/speed, Speed= distance/speed, and Distance= (speed)(time). we should memorize these.

Week 14 blog post

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On this week, we learned Rational Expressions and Equations. In these chapter, there are some new vocabulary. New words are rational expression, equivalent rational expressions, non permissible value rational equation, and rational equation. We have to know these words. In these words, the most important word is non permissible value. Non permissible value is rational expressions are not defined for values of the variable that make the denominator 0. These values are called non permissible values. Also in these chapter, we have to factor and find non permissible values.

Week 13 blog post

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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 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

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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

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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

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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 9 blog post

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On this week , we finished lesson 4.5 -4.7 and we reviewed midterm.

On these lessons, we learned how to change the equation to completing the square from y= ax^2+ bx +c, a > 0, we learned to find x-intercept by to factor. And we learned to graph a quadratic function from its equation in general form, also we learned to use a quadratic function to model a situation.

Ex) Write each equation in standard form

y =x^2 + 4x +8                                           y= 2x^2 +16x -10

y= (x^2 +4x +4-4)+8                                y= 2(x^2 +8x+16-16)-10

y=(x+2)^2 -4+8                                         y= 2(x+4)^2-32-10

y= (x+2)^2 +4                                            y= 2(x+4)^2-42

Ex)  Identify the x-intercepts of the graph of each quadratic function.

a) y= (x-12)(x+11)                                          b) y= (x+5)(x-17)

x-intercept : 12, -11                                    x- intercept : -5, 17

c) y=(2x+2)(x-6)                                            d) y= x(x-9)

x-intercept : -1, 6                                            x-intercept : 0,9

Ex) Determine the intercepts, the equation of the axis of symmetry, and the coordinates of the vertex of the graph of each quadratic function.

a) y= $latex 2x^2 +16x+24

y= 2(x^2 +8x+12)

y= $latex 2(x+6)(x+2)

x-intercept : -6, -2

y-intercept : 24

y= 2(x^2 + 8x+16-16)+12

y= $latex 2(x+4)^2-32+12

y= $latex 2(x+4)^2-20

Vertex : (-4, -20)

axis of symmetry : x=-4

Ex) A rectangular area is divided into 2 rectangles with 200 m of fencing used for the perimeter and the divider. What are the dimensions of the total maximum area.

3W+2L=200

2W/3 + L = 100

L = 100 – 2W/3

 

W * L = Maximum

 

W(100-2W/3)=Maximum

100W-2W^2/3 = Maximum

-2/3($latex W^2 – 200/3+40000/36 -40000/36)=Maximum

-2/3(W-200/6)^2+ 80000/108= Maximum

-2/3(W-200/6)^2+20000/27= Maximum

W = 100/3m

L = 50m

 

 

 

 

 

Week 8 blog post

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On this week, we learned new chapter. The new chapter is Analyzing Quadratic Functions. In this chapter, we learned how to draw the graph. The basic equation is y= a(x-p)^2 +q. In this equation, a means slope and if a is positive, it opens up and if it is negative, it opens down. Also when the graph is opens up, it has minimum vertex, and when the graph is opens down, it has maximum vertex. Also if a is 1, it always congruent with y= x^2. The next thing, p means horizontal translation, q means vertical translation. And x means x-intercept, y means y-intercept. And also if  absolute value a is bigger than zero, it is stretch and absolute value a is less than zero, it is compression. If we know these things, we can know how to be transformed at y= x^2.

Ex) The graph of y= x^2 is translated as described below. Without graphing, write the equation of the graph in its new position.

a) a translation of 16 units up and 13 units right                                b) a translation of 33 units down and 29 units left.

y= (x-13)^2 + 16                                                                  y= (x+29)^2 – 33

Also we have to know about the axis of symmetry. The axis of symmetry is a line through a shape so that each side is a mirror image. When the shape is folded in half along the axis of symmetry, then the two halves match up.

Ex) To find the axis of symmetry

a) y= (x-19)^2                   b) y= (x+7)^2                   c) y= x^2 +9

x=19                                                            x=-7                                                    x=0

EX) To find vertex

a) y= 5(x+7)^2 -13                    b) y= 0.164(x-26)^2+3

(-7,-13)                                                             (26,3)

 

 

 

 

Week 7 blog post

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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.