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

#### trip20

##### New member
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.

#### galactus

##### Super Moderator
Staff member
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

##### New member
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

##### Senior Member
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

##### Super Moderator
Staff member
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

##### New member
Thank you Glactus,

Fractions always confuse me.

#### galactus

##### Super Moderator
Staff member
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

##### New member
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

##### Super Moderator
Staff member
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

##### New member
DUH !!!!

I'd be dangerous if I could add.

Thank you Galactus,

EOM