Where a,b and c are constants
2006-11-30 13:47:20 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You have to complete the square.
y = a(x^2 + b/a x) + c
y = a(x^2 +b/a x + b^2/4a^2) + c - b^2/4a
y = a(x + b/2a)^2 + (4ac - b^2)/4a
[4ay - (4ac - b^2)]/4a = a(x + b/2a)^2
[4ay - 4ac + b^2]/(4a^2) = (x + b/2a)^2
+/- sqrt(4ay - 4ac + b^2)/2a = x + b/2a
x = (-b +/- sqrt(4ay - 4ac + b^2))/2a
2006-11-30 13:53:43 · answer #1 · answered by Puggy 7 · 0⤊ 0⤋
I'm not sure what the question is asking...if you need to find x-intercepts, and you have numbers for a, b, and c, you can factor and solve for x. Otherwise, you can use the equation x = [-b +/- sqrt(b^2 - 4ac) ] / 2a
Hope this helps!
2006-11-30 21:57:10 · answer #2 · answered by thingstealer 2 · 0⤊ 0⤋
Make this ax^2 + bx + (c - y) = 0
Then use the quadratic formula.
x = [-b +- sqrt (b^2 - 4a(c-y))] / 2a
2006-11-30 21:55:26 · answer #3 · answered by z_o_r_r_o 6 · 1⤊ 0⤋
You have to complete the square. This should lead you to the quadratic formula:x=-b+/- sq.rt.(b^2-4ac)/2a
2006-11-30 21:55:20 · answer #4 · answered by MateoFalcone 4 · 0⤊ 0⤋
Factor and set each paranthesis equal to zero
2006-11-30 21:51:41 · answer #5 · answered by sur2124 4 · 0⤊ 0⤋