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by **honest_denverco09** on Sat Apr 11, 2009 7:11 pm

We have x-intercept of graph is 4.8 , y-intercept is (- 5.76)
Find the quadratic function in general form ?
Thanks!

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by **stapel_eliz** on Sat Apr 11, 2009 7:37 pm

With only two points, you cannot determine a unique quadratic.

You'll have to make do with plugging (x, y) = (4.8, 0) and (x, y) = (0, -5.76) into y = ax² + bx + c, and do the best you can. You can clearly find the value of c, but you'll have to solve for a in terms of b, or vice versa.

by **honest_denverco09** on Sat Apr 11, 2009 8:06 pm

Thank you for your response!

But i did try that way before i posted this topic, and eventually, it can not be solved. I don't know why? Can you figure out where my mistake is, please ! Thanks.

I substituted 2 points (4.8 , 0) and (0 , -5.76) into the equation : " ax^2 + bx = 5.76 ", and i had the system : a(4.8)^2 + 4.8b = 5.76 and a(0) + b(0) = 5.76
And my result is "Math Error" ^^!

by **stapel_eliz** on Sat Apr 11, 2009 8:12 pm

I don't know what you mean by "Math Error"...? Also, what you have posted is not what was described. You seem instead to be working with y = ax² + bx, rather than y = ax² + bx + c, and to be using the points (0, +5.76) and (4.8, +5.76)...?

by **honest_denverco09** on Sat Apr 11, 2009 8:26 pm

"Math Error" means i can't solve this system of equation.

Because you advised me to substitute two points (4.8 , 0) and (0, -5.76) into y = ax^2 +bx + c , and you said that we have c = -5.76 . That is why i rewrote it into y = ax^2 + bx - 5.76 . And then i substituted them into this equation, i had the system : 0 = a(4.8)^2 + 4.8b - 5.76 or a(4.8)^2 + 4.8b = 5.76 and 5.76 = 0a +0b - 5.76 . So, the second equation in this system is unfounded. That is because i said "Math error" .

by **honest_denverco09** on Sat Apr 11, 2009 8:30 pm

Or did you mean i should substitute these two points into the equation y = ax^2 +bx + c and then solve for c ? If so, i can not solve for c with this system of equation anyway! :confused:

by **stapel_eliz** on Sat Apr 11, 2009 8:59 pm

Yes, you can find the value of "c". I've already specified how. :wink:

As for solving the system of equations, that's what I said at the outset: given only two points, you can not find numerical values for "the" quadratic. The best you can do is solve for one of "a" and "b" in terms of the other.

by **honest_denverco09** on Sat Apr 11, 2009 9:19 pm

Ok. i knew what you meant. But in this case, my teacher said, the answer is y = x^2 + 3.6x - 5.76 . i got more confused.

by **stapel_eliz** on Sat Apr 11, 2009 10:34 pm

Perhaps you were supposed to assume that the leading coefficient was 1...? Because there's no way to get that quadratic as "the" answer otherwise! :shock:

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