- #1
catwalk
- 4
- 0
I know this is the equation I have to use a(x - h) 2 + k, but don't know what points to use and how to covert to vertex form
In summary, the easiest way to find a, h, and k values for a quadratic equation is to choose three data points and do a google search for an online quadratic regression calculator.
I know this is the equation I have to use a(x - h) 2 + k, but don't know what points to use and how to covert to vertex form
- #2
HOI
- 921
- 2
Do you know what "best fit" means? There are several different types of "best fit". Which have you learned?
- #3
catwalk
- 4
- 0
No, I am not even sure what best fit means. I guess whatever is the simplestCountry Boy said:
- #4
HOI
- 921
- 2
Where did you get this problem? Did some evil teacher assign a problem about "best fit" without giving any instruction related to that?
(Don't you just hate when they do that?)
Have you tried looking up "best fit" in your textbook or online?
(Don't you just hate when they do that?)
Have you tried looking up "best fit" in your textbook or online?
- #5
catwalk
- 4
- 0
Haha yeah that's what happened.Country Boy said:
I've looked online but still don't know what to do
- #6
skeeter
- 1,103
- 1
The x-values of each data point seem to be easy ... 0, 2, 4, 5, 7, 9, 10, 12, and 14
You’ll just have to make your best eyeball estimate for the y-values.
Once you get a reasonable list of coordinates, do a google search for an online quadratic regression calculator and see what that can do for you.
You’ll just have to make your best eyeball estimate for the y-values.
Once you get a reasonable list of coordinates, do a google search for an online quadratic regression calculator and see what that can do for you.
- #7
HOI
- 921
- 2
Since you say you want a quadratic of the form a(x-h)^2+ k (notice the "^" for the exponent. That's clearer than just "x2".) You need to determine a, h, and k, three numbers. For that, you need three equations. The simplest thing to do is to choose three "data points" to get those equations. I recommend three widely spaced points so as not to exagerate any "anomaly" (like those two successive points with almost the same y which cannot happen in parabola). To the far left we have (0, 0), in the center, (7, 350), and on the right, (14, 80). In order that the parabola pass through those points we must have a(0- h)^2+ k= ah^2+k= 150, a(7- h)^2+ k= 350, and a(14-h)^2+ k= 80. Solve those equations for a, h, and k. Subtracting the first and third equations from the second eliminates "k" giving a(7- h)^2- ah^2= 200 and a(7-h)^2- a(14- h)^2= 270. Calculating the squares we have 49a- 14ha+ ah^2- ah^2= (49- 14h)a= 200 and 49a- 14ah+ ah^2- 196a+ 28ah- ah^2= a(14h- 147)= 270. Divide the second equation by the first to eliminate a, (14h- 147)/(49- 14h)= 270/200 and solve that equation for h.
A more "sophisticated" method would the "least squares" method which is harder but uses all of the data. With y= a(x-h)^2+ k, for each data point (xi, yi) the "error" is yi- a(xi- h)^2- k, the difference of the data value and the computed value. For example, for the first ^data point, (0, 0) the "error" is 0- a(0- h)^2- k= -ah^2- k. For (7, 350) the "error" is 350- a(7- h)^2- k. If we were to simply add those we might have a negative error cancelling a positive error that we do not want to happen. So square each and then sum. To find the smallest possible error, take the derivative of that sum of squares with respect to a, h, and k and set the derivatives equal to 0. Again that gives three equations to solve for a, h, and k.
A more "sophisticated" method would the "least squares" method which is harder but uses all of the data. With y= a(x-h)^2+ k, for each data point (xi, yi) the "error" is yi- a(xi- h)^2- k, the difference of the data value and the computed value. For example, for the first ^data point, (0, 0) the "error" is 0- a(0- h)^2- k= -ah^2- k. For (7, 350) the "error" is 350- a(7- h)^2- k. If we were to simply add those we might have a negative error cancelling a positive error that we do not want to happen. So square each and then sum. To find the smallest possible error, take the derivative of that sum of squares with respect to a, h, and k and set the derivatives equal to 0. Again that gives three equations to solve for a, h, and k.
FAQ: Solve a Quadratic Equation with Vertex Form
What is the vertex form of a quadratic equation?
The vertex form of a quadratic equation is y = a(x-h)^2 + k, where (h,k) is the coordinates of the vertex and a is the coefficient of the squared term.
How do I find the vertex of a quadratic equation in vertex form?
To find the vertex of a quadratic equation in vertex form, simply identify the values of h and k in the equation y = a(x-h)^2 + k. The vertex will be at the point (h,k).
Can you solve a quadratic equation with vertex form without factoring?
Yes, you can solve a quadratic equation with vertex form without factoring by using the quadratic formula: x = (-b ± √(b^2-4ac)) / 2a. Simply plug in the values of a, b, and c from your equation into the formula to find the solutions.
What is the significance of the a-value in the vertex form of a quadratic equation?
The a-value in the vertex form of a quadratic equation determines the direction and width of the parabola. If a is positive, the parabola opens upwards and is wider. If a is negative, the parabola opens downwards and is narrower.
Is it possible for a quadratic equation in vertex form to have no real solutions?
Yes, it is possible for a quadratic equation in vertex form to have no real solutions if the value inside the square root in the quadratic formula is negative. This indicates that the parabola does not intersect with the x-axis, and therefore has no real solutions.