Graphing F(x) = 2x^2 - 7x - 4

trip20

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Hello all,

I need to graph F(x) = 2x^2 - 7x - 4

I know that the Y intercept is (0, -4)

I think the X intercepts are -1 and 8

I'm really lost on the vertex. I ended up with (7/4, -10 1/4)

Any guidance would be GREATLY appreciated.

Thanks in advance!
 

galactus

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For the x intercepts, set to 0 and solve for x or just factor.

\(\displaystyle (x-4)(2x+1)=0\)

You have the y-intercept.

For the vertex coordinates:

Use \(\displaystyle x=\frac{-b}{2a}\). That will give you the x-coordinate of the vertex. y will follow.
 

trip20

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Thanks Galactus,

OK, so X intercepts are 4 and -1/2.

For the x vertex, -b/2a would equal 7/4, correct?

Then I get really lost when I plug that in to the equation for the y vertex

F(7/4)=2(7/4)^2-7(7/4)-4

I have no idea how to solve that?
 

Mrspi

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trip20 said:
Thanks Galactus,

OK, so X intercepts are 4 and -1/2.

For the x vertex, -b/2a would equal 7/4, correct?

Then I get really lost when I plug that in to the equation for the y vertex

F(7/4)=2(7/4)^2-7(7/4)-4

I have no idea how to solve that?
It's just arithmetic from this point. Follow the order of operations.

Do the powers first:

F(7/4) = 2(7/4)(7/4) - 2(7/4) - 4

F(7/4) = 2(49/16) - 2(7/4) - 4

Now, do the multiplications:

F(7/4) = (49/8) - (7/2) - 4

Now, do the additions and subtractions....remember that you will need a common denominator:

F(7/4) = (49/8) - (28/8) - (32/8)

F(7/4) = -11/8

So the coordinates of the vertex are (7/4, -11/8)
 

galactus

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Sorry, Mrspi, I believe that's

\(\displaystyle \L\\2(\frac{7}{4})^{2}-7(\frac{7}{4})-4=\frac{49}{8}-\frac{49}{4}-4=\frac{-81}{8}\)
 

trip20

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Thank you Glactus,

Fractions always confuse me.
 

galactus

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Mrspi done the work. I just pointed out a typo. Happens to us all.

Fractions needn't be a hindrance.

If we have: \(\displaystyle \frac{49}{8}-\frac{49}{4}-\frac{4}{1}\)

We have to make the denominators the same so we can subtract.

Use whatever you have to in order to make the denominator the largest one, in this case, 8:

\(\displaystyle \frac{49}{8}-\frac{49}{4}\cdot\frac{2}{2}-\frac{4}{1}\cdot\frac{8}{8}\)

\(\displaystyle \frac{49}{8}-\frac{98}{8}-\frac{32}{8}\)

\(\displaystyle \frac{49-98-32}{8}=\frac{-81}{8}\)

See?. Piece of cake....er...uh.....pi
 

trip20

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Galactus,

I'm really sorry to beat a dead horse but ....

wouldn't [ 2(7/4)^2 = 2(49/16) = (96/16) ?

giving us (96/16) - (49/4 * 4/4) - (4/1 * 16/16)

= (96/16) - (196/16) - (64/16)

=(-164/16) = (-82/8) = (-41/2)


thus, the vertex will be (7/4, -41/2)
 

galactus

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trip20 said:
Galactus,

I'm really sorry to beat a dead horse but ....

wouldn't [ 2(7/4)^2 = 2(49/16) = (96/16) ?

No, because 49*2=98, not 96.

giving us (96/16) - (49/4 * 4/4) - (4/1 * 16/16)

= (96/16) - (196/16) - (64/16)

=(-164/16) = (-82/8) = (-41/2)


thus, the vertex will be (7/4, -41/2)
\(\displaystyle \text{trust me, the vertex is}\)\(\displaystyle (\frac{7}{4},\frac{-81}{8})\)
 

trip20

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DUH !!!!

I'd be dangerous if I could add.

Thank you Galactus,

EOM
 
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