SOLUTION: Please help, We included the table of numbers that we have chosen. Any help, especially with graphing the answer would be highly appreciated. Thank you. When using the quadrat

Click here to see ALL problems on Quadratic Equations

Question 35931: Please help, We included the table of numbers that we have chosen. Any help, especially with graphing the answer would be highly appreciated. Thank you.

When using the quadratic formula to solve a quadratic equation ax2 + bx + c = 0, the discriminate is b2 - 4ac. This discriminate 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 discriminate means to the graph of y = ax2 + bx + c. Hint: Chose values of a, b and c to create a particular discriminate. Then, graph the corresponding equation.
DISCRIMINATE: b^2-4ac
A B C Discriminate
b^2-4ac
2 6 4 4
9 6 3 144
1 5 3 13
4 10 7 -12 {-3.4}
Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website!
SEE THE FOLLOWING AND TRY
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

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.

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...

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...