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Given f(x) = 11x^2 + 13x - 28, use the discriminant to determine whether or not f(x) ever equals -100 for any values of x. 

Best Answer 

 #2
avatar+93289 
+10

If you added 100 to f(x) then you would know.

$$\\11x^2+13x-28+100=0\\
11x^2+13x+72=0\\
\triangle=b^2-4ac\\
\triangle=13^2-4*11*72\\
\triangle<0\\
$therefore the graph will never reach as low as $ f(x)=-100$$

click here for the graph. 

Melody  Jan 29, 2015
 #1
avatar+88775 
+5

The discriminant = b^2 - 4ac.... so we have

13^2 - 4(11)(-28) = > 0    so this tells us we have real roots

But...maybe I'm missing something.....the discriminant doesn't tell us anything about whether or not f(x) ever = -100....it just tells us whether or not we may have "real" zeroes

We'd need to find the vertex for this

The x coordinate for the vertex is given by -b/2a  = -13/2(11)  = -13/22

And putting this back into the function.....we have the y value at the vertex = about -31.84

And since this parabola turns "up," (-13/22, -31.84) is the lowest point on the graph.....so f(x) never = -100....

 

CPhill  Jan 29, 2015
 #2
avatar+93289 
+10
Best Answer

If you added 100 to f(x) then you would know.

$$\\11x^2+13x-28+100=0\\
11x^2+13x+72=0\\
\triangle=b^2-4ac\\
\triangle=13^2-4*11*72\\
\triangle<0\\
$therefore the graph will never reach as low as $ f(x)=-100$$

click here for the graph. 

Melody  Jan 29, 2015
 #3
avatar+88775 
+5

Ah...that's very clever, Melody.....that thought didn't strike me...!!!

 

CPhill  Jan 29, 2015
 #4
avatar+93289 
+5

You often think of things that don't occur to me.  

One thing I like so much about this forum is that we can share little bits of info like this :)

the sharing makes learning so much easier :)

Melody  Jan 29, 2015

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