Question

What is the vertex of the function f(x)=2x^2+4x-16.
I think you start with (-b/2a, f (-b/2a))

Answer 1

@raindance, seems like a good start to me.

Answer 2

Have you attempted to figure out the vertex yet?

Answer 3

Yes. I got -1,-18

Answer 4

Would you mind showing your work please?

Answer 5

-4-2*2 = -1
2*-1^2-4-16=-18

Answer 6

\[-\frac{b}{2a} = -\frac{4}{2(2)} = -1\]
\[f(-\frac{b}{2a}) = f(-1) = 2(-1)^2 + 4(-1) - 16 = 2 - 4 - 16 = -18\]

Correct

Answer 7

I have one more question. Can you explain how to find the x-intercepts of f(x)=2x^2-10x+12.

Answer 8

@Hero

Answer 9

The x-intercepts of the graph of f(x) = 2x^2 - 10x + 12 occur where f(x) = 0. So...
Set f(x) = 0, then solve for x:


0 = 2x^2 - 10x + 12

Answer 10

In other words, solve the quadratic equation for x.

Answer 11

-12=2X^2-10X
I'm not sure hwat to do with the square...

Answer 12

Actually, you should begin by factoring out the 2 on the right side:

0 = 2(x^2 - 5x + 6)

Then divide both sides by 2:
0 = x^2 - 5x + 6

Now you have to factor x^2 - 5x + 6

Answer 13

I'm having a bit of a bran freeze is that finding two numbers that add to five and multiply to six?

Answer 14

That's part of the factoring process, yes :) Find two numbers that add to get -5 and multiply to get 6.

Answer 15

-2 and -3

Answer 16

For the next steps,

You could replace 5 with 2 + 3 as follows:

0 = x^2 - (2 + 3)x + 6

Then distribute -x as follows:
0 = x^2 -2x -3x + 6

Then factor the first two terms and the last two terms on the right side:
0 = x(x - 2) -3(x - 2)

And finally, factor out the remaining common term (x - 2):
0 = (x - 2)(x - 3)

Answer 17

Afterwards, you would use the zero product property to continue solving for x.

Answer 18

x-2=0 x=2
x-3=0 x=3