Rewrite c(x) = x^2 + 5x + 2 / x - 3 in the form y = p (x) + r(x)/b(x) where r has degree less than b
1 answer:
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3.) An extreme value refers to a point on the graph that is possibly a maximum or minimum. At these points, the instantaneous rate of change (slope) of the graph is 0 because the line tangent to the point is horizontal. We can find the rate of change by taking the derivative of the function.
y' = 2ax + b
Now that we where the derivative, we can set it equal to 0.
2ax + b = 0
We also know that at the extreme value, x = -1/2. We can plug that in as well.
The 2 and one-half cancel each other out.
Now we know that a and b are the same number, and that ax^2 + bx + 10 = 0 at x = -1/2. So let's plug -1/2 in for x in the original function, and solve for a/b.
a(-0.5)^2 + a(-0.5) + 10 = 0
0.25a - 0.5a + 10 = 0
-0.25a = -10
a = 40
b = 40
To determine if the extrema is a minima or maxima, we need to go back to the derivative and plug in a/b.
80x + 40
Our critical number is x = -1/2. We need to plug a number that is less than -1/2 and a number that is greater than -1/2 into the derivative.
LESS THAN:
80(-1) + 40 = -40
GREATER THAN:
80(0) + 40 = 40
The rate of change of the graph changes from negative to positive at x = -1/2, therefore the extreme value is a minimum.
4.) If the quadratic function is symmetrical about x = 3, that means that the minimum or maximum must be at x = 3.
y' = 2ax + 1
2a(3) + 1 = 0
6a = -1
a = -1/6
So now plug the a value and x=3 into the original function to find the extreme value.
(-1/6)(3)^2 + 3 + 3 = 4.5
The extreme value is 4.5
y' = 2ax + b
Now that we where the derivative, we can set it equal to 0.
2ax + b = 0
We also know that at the extreme value, x = -1/2. We can plug that in as well.
The 2 and one-half cancel each other out.
Now we know that a and b are the same number, and that ax^2 + bx + 10 = 0 at x = -1/2. So let's plug -1/2 in for x in the original function, and solve for a/b.
a(-0.5)^2 + a(-0.5) + 10 = 0
0.25a - 0.5a + 10 = 0
-0.25a = -10
a = 40
b = 40
To determine if the extrema is a minima or maxima, we need to go back to the derivative and plug in a/b.
80x + 40
Our critical number is x = -1/2. We need to plug a number that is less than -1/2 and a number that is greater than -1/2 into the derivative.
LESS THAN:
80(-1) + 40 = -40
GREATER THAN:
80(0) + 40 = 40
The rate of change of the graph changes from negative to positive at x = -1/2, therefore the extreme value is a minimum.
4.) If the quadratic function is symmetrical about x = 3, that means that the minimum or maximum must be at x = 3.
y' = 2ax + 1
2a(3) + 1 = 0
6a = -1
a = -1/6
So now plug the a value and x=3 into the original function to find the extreme value.
(-1/6)(3)^2 + 3 + 3 = 4.5
The extreme value is 4.5
Answer:
20%
Step-by-step explanation:
We can start by finding out how much the price of the television was reduced by.
250 - 200 = 50
Now, we need to know what percent of 250 is 50? Let x represent the percent of 250 that is 50. Using this, we can set up an equation:
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Notice that a percentage can also be written as that number over 100.
Reduce the fraction (250 and 100 have a common factor of 10)
Reduce the fraction again (25 and 10 have a common factor of 5)
Multiply both sides by two
Divide both sides by 5
Therefore the price of the television was reduced by 20%.