Answer:
f(x) = 4x^2 + 2x - 4.
Step-by-step explanation:
Let the quadratic function be y = f(x) = ax^2 + bx + c.
For the point (-2, 8) ( x = -2 when y = 8) we have:
a(-2)^2 + (-2)b + c = 8
4a - 2b + c = 8 For (0, -4) we have:
0 + 0 + c = -4 so c = -4. For (4, 68) we have:
16a + 4b + c = 68
So we have 2 systems of equations in a and b ( plugging in c = -4):
4a - 2b - 4 = 8
16a + 4b - 4 = 68
4a - 2b = 12
16a + 4b = 72 Multiplying 4a - 2b = 12 by 2 we get:
8a - 4b = 24
Adding the last 2 equations:
24a = 96
a = 4
Now plugging a = 4 and c = -4 in the first equation:
4(4) - 2b - 4 = 8
-2b = 8 - 16 + 4 = -4
b = 2.
The median is 38.5
The lower quartile is 29
There is an outlier (154)
The upper quartile is 70
and
The interquartile range is 70 - 29 = 41
It is proportional because we increase y.
I believe it’s c good luck!
Answer:
This question is solved in detail below. Please refer to the attachment for better understanding of an Ellipse.
Step-by-step explanation:
In this question, there is a spelling mistake. This is vertices not verticles.
So, I have attached a diagram of an ellipse in which it is clearly mentioned where are the vertices of an ellipse.
Vertices of an Ellipse: There are two axes in any ellipse, one is called major axis and other is called minor axis. Where, minor is the shorter axis and major axis is the longer one. The places or points where major axis and minor axis ends are called the vertices of an ellipse. Please refer to the attachment for further clarification.
Equations of an ellipse in its standard form:
This is the case when major axis the longer one is on the x-axis centered at an origin.
This is the case when major axis the longer one is on the y-axis centered at an origin.
where major axis length = 2a
and minor axis length = 2b