Question

Algebra II question, please see attachment!

Answer 1

I don't see any instructions for the problem on the attachment.

Answer 2

Are those supposed to be roots of a problem? If so, just FOIL them together

Answer 3

Find a quadratic equation in standard form for each set of points @Directrix

Answer 4

i think it's either the first or second equation

Answer 5

If x = 0, y has to be +4

Look at the options.

You can rule out the last two because of the +4 that results from substituting 0 in for x.

Answer 6

Look at the first two options and select one that we can use to test the given points to see if all of them are points of the function.

Answer 7

ohh okay

Answer 8

Let's test one of the top two. Which one?

Answer 9

try the first equation?

Answer 10

y = - x² + 5x - 4

We have to test this first for the point (0.4)

Answer 11

plug in 0 in the place of the x's?

Answer 12

Plug in the x for each equation and whichever one equals that x's y is correct

Answer 13

Yes.

Answer 14

I'll do the first point and you can do the other two if the first point fits the curve.

Answer 15

y = - x² + 5x - 4

Testing (0-,4)

y = - (0) ² + 5*0- 4

y = 0 +0 - 4 = -4

So far, so good.

It is your turn to test

y = - x² + 5x - 4 for the point (1,0), okay?

Answer 16

\[f(x)=-1+5-4\]
for (1,0)

Answer 17

or wait y = 1

Answer 18

What is -1 + 5 - 4 =

Answer 19

0

Answer 20

Correct. Now to check the last point.

Answer 21

\[-(2)^2 + 5(2) - 4 = y = 2\]
for (2,2)

Answer 22

Correct. We have found the answer to the question.

Answer 23

which one would it be? like how do you make it in that form

Answer 24

This is the function we tested:

f(x)= - x² + 5x - 4

Answer 25

Oh okay, brain fart lol

Answer 26

@CloverRacer Question?

Answer 27

No question, thank you for the explanation!

Answer 28

You are welcome. Thanks for participating in the discussion.