- #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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

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.

- #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

Mathematics news on Phys.org

- #2

HOI

- 921

- 2

- #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

(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:

(Don't you just hate when they do that?)

Have you tried looking up "best fit" in your textbook or online?

I've looked online but still don't know what to do

- #6

skeeter

- 1,103

- 1

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

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.

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.

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).

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.

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.

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.

- Replies
- 44

- Views
- 3K

- Replies
- 7

- Views
- 3K

- Replies
- 3

- Views
- 2K

- Replies
- 1

- Views
- 1K

- Replies
- 1

- Views
- 1K

- Replies
- 1

- Views
- 1K

- Replies
- 3

- Views
- 1K

- Replies
- 2

- Views
- 4K

- Replies
- 7

- Views
- 2K

- Replies
- 5

- Views
- 1K

Share: