por Kassandra Perez hace 6 meses
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The height of an object launched upward, ignoring the effects of air resistance, can be modeled with the following formula:
height=−12gt2+v0t+s0
Using function notation, which is more appropriate, we have
With this formula, the height can be calculated at any given time t after the object is launched. The coefficients represent the following:
−12g The letter g represents the acceleration due to gravity.
v0 “ v -naught” represents the initial velocity of the object.
s0 “ s -naught” represents the initial height from which the object is launched
We consider only problems where the acceleration due to gravity can be expressed as g=32 ft/sec 2 . Therefore, in this section time will be measured in seconds and the height in feet. Certainly though, the formula is valid using units other than these.
Find the vertex of the quadratic equation.
Determine the y -value of the vertex.
The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers.
Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?
we knew the number of subscribers to a newspaper and used that information to find the optimal price for each subscription. What if the price of subscriptions is affected by competition?
Previously, we found a quadratic function that modeled revenue as a function of price. Revenue− 2,500p2+159,000p We found that selling the paper at $31.80 per subscription would maximize revenue. What if your closest competitor sells their paper for $25.00 per subscription? What is the maximum revenue you can make you sell your paper for the same?
So the area of the rectangular region is length times width, which in terms of the unknown w is (2w+3)w or 2w2+3w
Now, this looks like a quadratic equation, at least on the surface, but we need to make sure it isn't a linear equation in disguise so we subtract 4w2 from both sides of the equation to get 0=−2w2+3w.
e will not show this in detail, but an equation of the form 0= (something quadratic) is not a linear equation. This is the assertion about the width in the problem! Whatever w is, it satisfies the equation 0=−2w2+3w. For example, w≠1 since if it were, −2w2+3w=−2(1)2+3(1)=1 which is not 0 as asserted. Our question is then, how do you find the width (so that if you substitute that width in for w the assertion is true! In general we will be interested in finding the solutions to quadratic equations.
If the length of the rectangular region is 3 more than twice its width, then its length is 2w+3
The area of a square region with width w is (2w)2 or, 4x2
2x+1 , 0 2x+1=0 (linear equation) x2 x−2 x2=x−2 (quadratic equation) 3x2+5x−1 0 3x2+5x−1=0 (quadratic equation)
Now, by the zero product property, either x+5=0 or x−4=0 which means either x=−5 or x=4 These are the two solutions of the equation.
You can factor the left side as: (x+5)(x−4)=0
Suppose you want to solve the equation x2+x−20=0
Imaginary Number Rules
Consider the pure quadratic equation: x 2 = a, where ‘a’ is a known value. Its solution may be presented as x = √a. Therefore, the rules for some imaginary numbers
i = √-1 i2 = -1 i3 = -i i4 = +1 i4n = 1 i4n-1= -i
Consider an example, a+bi is a complex number. For a +bi, the conjugate pair is a-bi. The complex roots exist in pairs so that when multiplied, it becomes equations with real coefficients.
Operations on Imaginary Numbers
Subtraction of Numbers Having Imaginary Numbers
When c+di is subtracted from a+bi, the answer is done like in addition. It means, grouping all the real terms separately and imaginary terms separately and doing simplification. Here, (a+bi)-(c+di) = (a-c) +i(b-d).
Addition of Numbers Having Imaginary Numbers
When two numbers, a+bi, and c+di are added, then the real parts are separately added and simplified, and then imaginary parts separately added and simplified. Here, the answer is (a+c) + i(b+d).
Multiplication of Numbers Having Imaginary Numbers
Consider (a+bi)(c+di) It becomes: (a+bi)(c+di) = (a+bi)c + (a+bi)di = ac+bci+adi+bdi2 = (ac-bd)+i(bc+ad)
Division of Numbers Having Imaginary Numbers
Consider the division of one imaginary number by another. (a+bi) / ( c+di) Multiply both the numerator and denominator by its conjugate pair, and make it real. So, it becomes (a+bi) / ( c+di) = (a+bi) (c-di) / ( c+di) (c-di) = [(ac+bd)+ i(bc-ad)] / c2 +d2.
Imaginary Numbers Example
Solve the imaginary number i7
The given imaginary number is i7 Now, split the imaginary number into terms, and it becomes i7 = i2 × i2 × i2 × i i7 = -1 × -1 × -1 × i i7 = -1 × i i7 = – i Therefore, i7 is – i. Keep visiting BYJU’S – The Learning App and also register with it to watch all the interactive videos.
The basic arithmetic operations in Mathematics are addition, subtraction, multiplication, and division. Let us discuss these operations on imaginary numbers. Let us assume the two complex numbers: a + bi and c + di
Complex numbers are the combination of both real numbers and imaginary numbers. The complex number is of the standard form: a + bi Where a and b are real numbers i is an imaginary unit. Real Numbers Examples : 3, 8, -2, 0, 10 Imaginary Number Examples: 3i, 7i, -2i, √i Complex Numbers Examples: 3 + 4 i, 7 – 13.6 i, 0 + 25 i = 25 i, 2 + i.
Imaginary numbers are an important mathematical concept;
imaginary numbers are numbers that are not real. We know that the quadratic equation is of the form ax2 + bx + c = 0, where the discriminant is b2 – 4ac. Whenever the discriminant is less than 0, finding square root becomes necessary for us. Here, we are going to discuss the definition of imaginary numbers, rules and its basic arithmetic operations
Imaginary Numbers Definition
Imaginary numbers are the numbers when squared it gives the negative result. In other words, imaginary numbers are defined as the square root of the negative numbers where it does not have a definite value. It is mostly written in the form of real numbers multiplied by the imaginary unit called “i”.
The general equation of a parabola is: y = a(x-h)2 + k or x = a(y-k)2 +h, where (h,k) denotes the vertex.
Formula:
The parabola’s Vertex Formula serves to determine the coordinates of the point where the parabola intersects its axis of symmetry, known as the vertex, denoted as (h, k). The standard equation for a parabola is represented as y = ax2 + bx + c. When the coefficient of x2 is positive, the vertex is positioned at the lowest point of the U-shaped curve, whereas a negative coefficient places the vertex at the highest point of the U-shaped curve.
The vertex corresponds to the minimum point when the parabola opens upward or the maximum point when it opens downward, marking the turning point where the parabola changes its direction. Exploring the vertex formula further and working through examples will enhance understanding of its application in solving parabolic equations.
The Vertex Formula is instrumental in determining the coordinates of the vertex for a parabola. The standard parabolic equation is expressed as y = ax2 + bx + c. In the vertex form of the parabola, it’s represented as y = a(x – h)2 + k. The vertex coordinates (h, k) can be found using two methods: (h,k)=(− 2a b ,− 4a D ), where D (the discriminant) = b 2 −4ac (h,k) can be found by setting h=− 2a b and then evaluating y at h to determine the value of
Derivation of Vertex Formulas
tarting with the standard form of a parabola, y = ax2 + bx + c, the conversion to the vertex form y = a(x – h)^2 + k is achieved by completing the square. First, by subtracting c from both sides, we get: y – c = ax2 + bx Factoring out ‘a’: y – c = a(x2 + b/a x) Identifying half of the coefficient of x as b/2a and its square as b2/4a2, we add and subtract this term within the parentheses on the right side: y – c = a(x2 + b/a x + b2/4a2 – b2/4a2)
This expression simplifies to: y – c = a((x + b/2a)2 – b2/4a2) E Expanding and rearranging terms, adding ‘c’ to both sides: y = a(x + b/2a)2 – b2/4a + c y = a(x + b/2a)2 – (b2 – 4ac) / (4a) Comparing this with the vertex form y = a(x – h)2 + k, the following vertex values are derived: h = -b/2a k = -(b2 – 4ac) / (4a)
This process leads to the vertex form y = a(x - h)^2 + k, where h = -b / (2a) and k = - (b^2 - 4ac) / (4a).
parabola
vertex of a parabola
maximum value
minimum value
vertex
axis of symmetry
Forms of Quadratic Functions
A quadratic function is a function of degree two. The graph of a quadratic function is a parabola. The general form of a quadratic function is f(x)=ax2+bx+c where a, b and are real numbers and a≠0 The standard form of a quadratic function is f(x)=a(x−h)2+k The vertex (h,k) is located at h=–b2a,k=f(h)=f(−b2a
where a b and c are real numbers and a≠0 If a>0 the parabola opens upward. If a<0 the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry.
The standard form of a quadratic function presents the function in the form
where (h,k) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.
The three major functional operations of an organization include marketing, finance, and operations.
General Form of a parabola is y = ax2 + bx + c. In order to find the vertex from this form, you must first find the x-coordinate of the vertex which is x = - b/2a. After you find the x-coordinate of the vertex, you will take this number and substitute for x in the parabola equation.
What is the formula for the vertex form of a parabola?
Vertex Form of Parabola. We know that the standard form of the parabola is y=ax2+bx+c. Thus, the vertex form of a parabola is y = a(x-h)2 + k.
Given the focus (h,k) and the directrix y=mx+b, the equation for a parabola is (y - mx - b)^2 / (m^2 +1) = (x - h)^2 + (y - k)^2.
Get the equation in the form y = ax2 + bx + c. Calculate -b / 2a. This is the x-coordinate of the vertex. To find the y-coordinate of the vertex, simply plug the value of -b / 2a into the equation for x and solve for y.
The vertex is the turning point of the graph. We can see that the vertex is at (3,1) ( 3 , 1 ) . The axis of symmetry is the vertical line that intersects the parabola at the vertex.
If you are given factored form you need to take the product of your a and b value. A positive value means the parabola opens up, and a negative value means the parabola opens down.
The axis of symmetry always passes through the vertex of the parabola . The -coordinate of the vertex is the equation of the axis of symmetry of the parabola. For a quadratic function in standard form, y = a x 2 + b x + c , the axis of symmetry is a vertical line x = − b 2 a
Subtovertex is the highest or lowest point on a parabola, its y-coordinate is the maximum value or minimum value of the function. The vertex of a parabola lies on the axis of the parabola. So, the graph of the function is increasing on one side of the axis and decreasing on the other sidepic
The zeros of a parabola are the points on the parabola that intersect the line y = 0 (the horizontal x-axis).
The discriminant formula is d=b^2-4ac given the equation of the quadratic is f(x)=ax^2+bx+c.
Multivariate
Polynomial
Grouping
f(x,y)=f(−x,−y)
Subtopic
Visually we have that the y-axis acts as a mirror for the graph. We will demonstrate several functions to test for symmetry graphically using the graphing calculator!
Symmetric with respect to the x-axis
f(x,y)=f(x,−y).
We say that a graph is symmetric with respect to the x axis if for every point (a,b) on the graph, there is also a point (a,−b) on the graph; hence
rotation
reflection
glide reflection
Determining Whether a Function is Even, Odd, or Neither
even functions are symmetrical about the y-axis: f(x)=f(-x). Odd functions are symmetrical about the x- and y-axis: f(x)=-f(-x).
Determine whether the function satisfies f(x)=f(−x) f ( x ) = f ( − x ) . If it does, it is even. Determine whether the function satisfies f(x)=−f(−x) f ( x ) = − f ( − x ) . If it does, it is odd. If the function does not satisfy either rule, it is neither even nor odd.
Evaluating Piecewise Function
piecewise-defined function is one which is defined not by a single equation, but by two or more.
A piecewise function is a function built from pieces of different functions over different intervals. For example, we can make a piecewise function f(x) where f(x) = -9 when -9 < x ≤ -5, f(x) = 6 when -5 < x ≤ -1, and f(x) = -7 when -1.
The basic idea behind piecewise linear regression is that if the data follow different linear trends over different regions of the data then we should model the regression function in "pieces."
summary. A piecewise function is a function where more than one formula is used to define the output over different pieces of the domain.
A piecewise function is a function that is defined on a sequence of intervals. A common example is the absolute value
How to graph the given piecewise-defined function?
The method for graphing piecewise functions involves first identifying the intervals the describe each of the subdomains. Then, correlate each subfunction with each of these intervals. Next, graph each of the subfunctions on their subdomains omitting any points that are not in the interval.
piecewise function is the absolute value function. The absolute value function can be written as f ( x ) = | x | . This function has three different pieces, x for all values less than 1, x 2 for all values between 1 and 3 (including 1 and 3) and finally, x + 3 for all values greater than three.
Increasing and Decreasing Functions
Decreasing Function - A function f(x) is said to be decreasing on an interval I if for any two numbers x and y in I such that x < y, we have f(x) ≥ f(y).
Increasing Function - A function f(x) is said to be increasing on an interval I if for any two numbers x and y in I such that x < y, we have f(x) ≤ f(y).
For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece.
Given a piecewise function, sketch a graph. Indicate on the x− axis the boundaries defined by the intervals on each piece of the domain.
To evaluate a piecewise function at any given input, first, see which of the given intervals (or inequalities) the given input belongs to. Then just substitute the given input in the function definition corresponding to that particular interval.
Algebraically, a function is even if for all possible values. For example, for the even function below, notice how the -axis symmetry ensures that for all
Algebraically, a function is odd if for all possible values. For example, for the odd function below, notice how the function's symmetry ensures that f(-x) is always the opposite of f(x)
For y=x3 we replace with (−y)=(−x)3 so that −y=−x3 or y=x3 Hence y=x3 is symmetric with respect to the origin.
Example
x−2y=5
we replace with
x−2(−y)=5
Which is not equivalent to the original expression. So
Is not symmetric with respect to the x-axis.
How do you explain increasing and decreasing functions?
Function f is increasing in [p, q] if f′(x) > 0 for each x ∈ (p, q). Function f is decreasing in [p, q] if f′(x) < 0 for each x ∈.
Determining the Intervals over Which a Function Increases, Decreases, or Constant with Closed Intervals
So if we want to find the intervals where a function increases or decreases, we take its derivative an analyze it to find where it's positive or negative
How can we tell if a function is increasing or decreasing?
If f′(x)<0 on an open interval, then f is decreasing on the interval.
a function is increasing on an interval if the function values increase as the input values increase within that interval.
Determining the Intervals over Which a Function Increases, Decreases, or is Constant with Open Intervals
The intervals where a function is increasing (or decreasing) correspond to the intervals where its derivative is positive (or negative). So if we want to find the intervals where a function increases or decreases, we take its derivative an analyze it to find where it's positive or negative
If f′(x)>0 on an open interval, then f is increasing on the interval.
The intervals where a function is increasing (or decreasing) correspond to the intervals where its derivative is positive (or negative).
