Question 2032: In a case of a quadratic equation, why are there two x-intercepts and one y-intercept?
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! If a given quadratic equation as:
y = ax^2 +b x + c. (Note a <> 0,& b,c reals)
To find x-intercept,since the corresponding point which y-coordinate
must be 0. We have to solve the equation
y = ax^2 +b x + c = 0...(*).
In general,if the discriminant b^2 - 4ac > 0, (*) has two different
real roots and hence there are two x-intercepts
(-b+sqrt(b^2 - 4ac))/2a or (-b-sqrt(b^2 - 4ac))/2a if b^2 - 4ac > 0.
[Note: IF b^2 - 4ac = 0 , there is only one x-intercept -b/2a
IF b^2 - 4ac < 0 , there are no x-intercepts ]
To find y-intercept,since the corresponding point which x-coordinate
must be 0. When we set x =0, we obtain
y = c. Hence, there is only one y-intercept c.
Kenny
|
|
|
