Question 36483
SEE THE FOLLOWING EXAMPLES AND COME BACK IF STILL IN DOUBT
When using the quadratic formula to solve a quadratic equation ax2 + bx + c = 0, the discriminant is b2 - 4ac. This discriminant can be positive, zero, or negative. (When the discriminate is negative, then we have the square root of a negative number. This is called an imaginary number, sqrt(-1) = i. )
Explain what the value of the discriminant means to the graph of y = ax2 + bx + c. Hint: Chose values of a, b and c to create a particular discriminant. Then, graph the corresponding equation.
CASE 1....DISCRIMINANT (D SAY) IS POSITIVE.....
EX..LET Y = X^2-5X+6=0..
D=5^2-4*1*6=25-24=1...
HENCE ROOTS ARE 
REAL,THAT IS THE GRAPH CUTS THE X AXIS AT 2 REAL POINTS
DISTINCT ...2 AND 3
AND THE FUNCTION Y COULD BE POSITIVE OR NEGATIVE WITH A MAXIMUM OR MINIMUM
SEE GRAPH BELOW
{{{ graph( 500, 500, -10, 10, -10, 10, x^2-5x+6) }}}
CASE 2.....D=0
EX....LET...Y=X^2-2X+1=0
D=2^2-4*1*1=4-4=0
HENCE ROOTS ARE 
REAL.THAT IS THE GRAPH CUTS THE X AXIS AT 1 REAL POINT.
EQUAL...1 AND 1 
AND THE FUNCTION Y IS ALWAYS NON NEGATIVE OR NON POSITIVE DEPENDING ON THE SIGN OF COEFFICIENT OF X^2 BEING POSITIVE OR NEGATIVE , WITH A MINIMUM OR MAXIMUM VALUE OF ZERO.
SEE GRAPH BELOW.
{{{ graph( 500, 500, -10, 10, -10, 10, x^2-2x+1) }}}
CASE 3.......D IS NEGATIVE
EX....LET Y = X^2+X+1=0
D=1^2-4*1*1=1-4=-3
HENCE ROOTS ARE
IMAGINARY.THAT IS THE GRAPH DOES NOT CUT  THE X AXIS.
DISTINCT.....(-1+iSQRT(3))/2....AND (-1-iSQRT(3))/2
AND THE FUNCTION  Y IS ALWAYS POSITIVE SINCE COEFFICIENT OF X^2 IS POSITIVE.
SEE THE GRAPH BELOW...
{{{ graph( 500, 500, -10, 10, -10, 10, x^2+x+1) }}}
 
EX....LET Y = -X^2+X-1=0
D=1^2-4*(-1)*(-1)=1-4=-3
HENCE ROOTS ARE
IMAGINARY.THAT IS THE GRAPH DOES NOT CUT THE X AXIS.
DISTINCT.....(1-iSQRT(3))/2....AND (1+iSQRT(3))/2
AND THE FUNCTION  Y IS ALWAYS NEGATIVE SINCE COEFFICIENT OF X^2 IS NEGATIVE.
SEE THE GRAPH BELOW...
{{{ graph( 500, 500, -10, 10, -10, 10, -x^2+x-1) }}}