Functions:using discriminant to find number of x-intercepts

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Homework Statement


upload_2016-5-21_13-0-5.png


Homework Equations


Discriminant: b2 - 4ac

The Attempt at a Solution


a) [/B]f(x) = 3x2 – 5x + 1

y = 3x2 – 5x +1

Substitute a=3, b= -5, c=1 into the discriminant:

b2 – 4ac = (-5)2 – 4(3)(1)

=25 – 12

= 13 This number is positive.

Since b2 – 4ac > 0, there are two real roots, so the quadrative function has two x-intercepts.

b) f(x) = 2x2 + x +1

y = 2x2 + x +1

Substitute a=2, b=1, c=1 into the discriminant:

b2 – 4ac = (1)2 – 4(2)(1)

= 1 – 8

= -7 This number is negative.

Since b2 – 4ac < 0, there are no real roots, so the quadratic function does not have any x-intercepts.

c) f(x) = 4x2 – 12x + 9

y = 4x2 – 12x + 9

Substitute a=4, b= -12, c=9 into the discriminant:

b2 – 4ac = (-12)2 – 4(4)(9)

=144 – 144

= 0

Since b2 – 4ac = 0, there is one (double) real root, so the quadratic function has one x-intercept.

Is this correct?
 
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This is absolutely correct. You can additionally verify by drawing the graphs for the equations.
 
Evangeline101 said:
Okay. Thanks for verifying my answer :biggrin:

You are welcome. However you could have done it yourself by hand-drawing the graph or using a graphing calculator.
 
Mastermind01 said:
However you could have done it yourself by hand-drawing the graph or using a graphing calculator.

Oh okay. But this is the way they taught me in the lesson, they did not ask me to graph anything.
 
Evangeline101 said:
Oh okay. But this is the way they taught me in the lesson, they did not ask me to graph anything.

What you did / they taught is correct. What I am saying is instead of asking here you could have verified your answer yourself by graphing. It's also a good way to check your answer when you don't have access to the forum.
 
Mastermind01 said:
What you did / they taught is correct. What I am saying is instead of asking here you could have verified your answer yourself by graphing. It's also a good way to check your answer when you don't have access to the forum.

Oh okay, I get it :)
Thanks again for the help.
 
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