You throw a rock off the balcony overlooking your backyard the table shows the height of the rock at different times use quadratic regressin to find a quadratic model for this data.
time
0 1 2 3 4 5
height
24 44.2 54.8 55.8 47.2 29
would someone explain this to me had a problem like this before and got it wrong
Judging from your other problems, I think quadratic regression is a bit beyond the current subject matter. (If not so, let me know just what you have studied in regards to quadratic regression.) So let's try a simpler approach.
evidently the equation is
y = ax^2 + bx + 24
plugging in the given values,
20.2 = a + b
30.8 = 4a + 2b
31.5 = 9a + 3b
23.2 = 16a + 4b
5 = 25a + 5b
Solving these equations in pairs, we get
(a,b) = (-4.8,25),(-4.9,25.2),(-4.7,24.6),(-4.8,25) and so forth.
If you're not looking for the best fit, you could use a model like
y = -4.8x^2 + 25x + 24
If this approach is not what you wanted, please clarify.
The y=4.8x^2+25x+24 was correct answer, but I'm not sure how you got that.
the actual answer to this is
y=-4.8^2+25x+24 you did not put a negative on it :)
1-7 answers for Factoring quadratic quiz
1. C
2. C
3. B
4. A
5. D
6. B
7. B
You're welcome now do good in school kids!
Suppose a parabola has vertex (-4,7). and also passes through the point (-3,8) write the equation in a a parabola vertex form
Sure, I can help you with that!
To find a quadratic model for the given data, we can first plot the data points on a graph. Here is the data plotted on a graph:
(0, 24)
(1, 44.2)
(2, 54.8)
(3, 55.8)
(4, 47.2)
(5, 29)
Now, we can try to fit a quadratic function of the form y = ax^2 + bx + c to these data points. Since we have six data points, we will have three unknowns (a, b, and c) and three equations.
The equation of a quadratic function is: y = ax^2 + bx + c
Plugging in the first data point (0, 24):
24 = a(0^2) + b(0) + c
24 = c
So, we now know that c = 24.
Now, let's find the equations using the remaining data points:
For the point (1, 44.2):
44.2 = a(1^2) + b(1) + 24
44.2 = a + b + 24
a + b = 20.2
For the point (2, 54.8):
54.8 = a(2^2) + b(2) + 24
54.8 = 4a + 2b + 24
4a + 2b = 30.8
2a + b = 15.4 (dividing both sides by 2)
For the point (3, 55.8):
55.8 = a(3^2) + b(3) + 24
55.8 = 9a + 3b + 24
9a + 3b = 31.8
3a + b = 10.6 (dividing both sides by 3)
Solving the system of equations:
a + b = 20.2 (Equation 1)
2a + b = 15.4 (Equation 2)
3a + b = 10.6 (Equation 3)
Subtracting Equation 1 from Equation 2:
a = -4.8
Substituting the value of a into Equation 1:
-4.8 + b = 20.2
b = 25
Now we have a = -4.8 and b = 25. Using c = 24, we can plug these values back into the quadratic equation:
y = -4.8x^2 + 25x + 24
Thus, the quadratic model for the given data is y = -4.8x^2 + 25x + 24.