Completing the square is method of solving a quadratic equation that involves finding a value to add to both the left and right side of the equation. This value has the extra benefit of making one side of the equation a perfect trinomial which makes the function easier to identify and/or graph.
Let's begin by factoring
Take the coefficient of the
We added these values but the equation remains balanced because they are added to both sides of the equation.
The sequence is a geometric sequence.
The sequence given is :
1) Checking if the sequence is an arithmetic sequence:
As observed
2) Checking if the sequence is a geometric sequence:
Since
So the sequence is a geometric sequence.
A piecewise continuous function is a function that is continuous except at a finite number of points in its domain.
Note that the points of discontinuity of a piecewise continuous function do not have to be removable discontinuities. That is we do not require that the function can be made continuous by redefining it at those points. It is sufficient that if we exclude those points from the domain, then the function is continuous on the restricted domain.
For example, consider the function:
graph{(y - x/abs(x))(x^2+y^2-0.001) = 0 [-5, 5, -2.5, 2.5]}
This is continuous for all
The discontinuity at
At
So the left limit and right limit disagree with one another and with the value of the function at
If we exclude the finite set of discontinuities from the domain, then the function restricted to this new domain will be continuous.
In our example, the definition of
If we graph
Slightly confusingly, the function
graph{tan(x) [-10.06, 9.94, -4.46, 5.54]}
Meanwhile, the sawtooth function
graph{3/5(abs(sin(x * pi/2))-abs(cos(x * pi/2))-abs(sin(x * pi/2)^3)/6+abs(cos(x * pi/2)^3)/6) * tan(x * pi/2)/abs(tan(x * pi/2))+1/2 [-2.56, 2.44, -0.71, 1.79]}
Via changing velocity I pressume you mean an object that accelerates or decelerates.
If acceleration is constant
If you have initial and final speed:
Usually
If the above method does not work because you are missing some values, you can use the equation below. The distance traveled
where
Therefore, if you know the distance, initial speed and acceleration you can find the time by solving the quadratic equation that is formed. However, if acceleration if not given, you will need the final speed of the object
and substitute to the distance equation, making it:
Factor
So you got 2 equations. Pick one of them, which will help you solve with the data you are given:
Below are two other cases where acceleration is not constant. FEEL FREE TO IGNORE THEM if acceleration in your case is constant, since you placed it in the Precalculus category and the below contain calculus.
If acceleration is a function of time
The definition of acceleration:
If you still don't have enough to solve, that means you have to go to distance. Just use the definition of speed and move on, as if I analyze it further it will only confuse you:
The second part of this equation means integrading acceleration with respect to time. Doing that gives an equation with only
If acceleration is a function of speed
The definition of acceleration:
The partial fraction decomposition suggests that the function can be broken down into the sum of two other functions, or;
Where we need to solve for
We can now cancel the denominator on each side, leaving;
Now we can solve for
The
Solving for
We can substitute
This time, the
Now that we have our values for
Look at changes of signs to find this has
Then do some sums...
#f(x) = -3x^4-5x^3-x^2-8x+4#
Since there is one change of sign,
#f(-x) = -3x^4+5x^3-x^2+8x+4#
Since there are three changes of sign
Since
Newton's method can be used to find approximate solutions.
Pick an initial approximation
Iterate using the formula:
#a_(i+1) = a_i - f(a_i)/(f'(a_i))#
Putting this into a spreadsheet and starting with
#x ~~ 0.41998457522194#
#x ~~ -2.19460208831628#
We can then divide
Notice the remainder
Check the discriminant of the approximate quotient polynomial:
#-3x^2+0.325x-4.343#
#Delta = b^2-4ac = 0.325^2-(4*-3*-4.343) = 0.105625 - 52.116 = -52.010375#
Since this is negative, this quadratic has no Real zeros and we can be confident that our original quartic has exactly
The general standard form for the equation of a circle is
for a circle with center
Given
We can transform this into the standard form by the following steps:
Move the
Complete the square for each of the
Re-write the
Often we would leave it in this form as "good enough",
but technically this wouldn't make the
So more accurately:
with center at
An exponential function is in the general form
#y=a(b)^x#
We know the points
#8=a(b^-1)=a/b#
#2=a(b^1)=ab#
Multiply both sides of the first equation by
#8b=a#
Plug this into the second equation and solve for
#2=(8b)b#
#2=8b^2#
#b^2=1/4#
#b=+-1/2#
Two equations seem to be possible here. Plug both values of
If
#2=a(1/2)#
#a=4#
Giving us the equation:
If
#2=a(-1/2)#
#a=-4#
Giving us the equation:
However! In an exponential function,
The only valid function is
#color(green)(y=4(1/2)^x#
Given that a point
Then
This is not a polar graph!!!
We are give the coordinates of (-4,3)
Suppose we viewed this in the context of Cartesian form and use
Then
So we would have
Suppose the graph was only plotted over the range
Then the above graph would not be continuous but be a line from
All we need now is the angle that that line makes to the x-axis and the length of that line.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The Polar angle
Let the angle from the line to the negative x-axis be
Then
But
so
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given that a point
Then
The first thing that has to be done is to factorise the denominator.
#a^3 - b^3 = (a - b )(a^2 + ab + b^2 ) #
here then
So
Now
#(8x - 1) /(x^3 - 1 )= A/(x - 1 ) +( Bx + C)/(x^2 + x + 1 )#
Multiplying through by
#8x - 1 = A(x^2 + x + 1 ) + (Bx + C )(x - 1 )" " " " color(red)(("*"))#
We now have to find the values of
Note that if we use
Substitute x = 1 in equation
#7 = 3A + 0 rArr A = 7/3#
To find B and C it will be necessary to compare the coefficients on both sides of the equation
#rArr 8x -1 = Ax^2 + Ax + A + Bx^2 + Cx - Bx - C #
This can be 'tidied up' by collecting like terms and letting
#8x - 1 = 7/3 x^2 + 7/3 x + 7/3 + Bx^2 + Cx - Bx - C #
Compare
#0 = 7/3 + B rArr B = -7/3 #
Now compare constant terms.
# - 1 = A - C = -7/3 - C rArr C= -10/3 #
Finally
#(8x - 1 )/(x^3 - 1 ) =( 7/3)/(x - 1 ) + ( -7/3x - 10/3)/(x^2 +x + 1 ) #