What is the greatest integer that satisfies the inequality below?
1 answer:
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The longest side
Think about it. The widest angle would leave a resulting long side to reach the ends of the angle
If you look at the picture, the side across from the widest angle has the longest length
Think about it. The widest angle would leave a resulting long side to reach the ends of the angle
If you look at the picture, the side across from the widest angle has the longest length
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