SOLUTION: Use the quadratic formula to solve a quadratic equation ax^2 + bx + c = 0, the discriminant is b^2-4ac. This discriminant can be positive, zero, or negative. Explain what the valu

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Use the quadratic formula to solve a quadratic equation ax^2 + bx + c = 0, the discriminant is b^2-4ac. This discriminant can be positive, zero, or negative. Explain what the valu      Log On


   

Question 36483: Use the quadratic formula to solve a quadratic equation ax^2 + bx + c = 0, the discriminant is b^2-4ac. This discriminant can be positive, zero, or negative. Explain what the value of the disciminant means to the graph of y= ax^2 + bx + c. Use these values of a=2, b=4,and c=5. Then, graph the corresponding equation.
How do I know what the discriminant is? If I use the quadratic equation, does this show how to graph the equation? I think the vertex is -1,4 but how do I know how to graph the rest of the parabola?
I really need some help with this one. Thanks
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
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%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2-5x%2B6%29+
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%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2-2x%2B1%29+
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%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2%2Bx%2B1%29+

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%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+-x%5E2%2Bx-1%29+