Question

given thanks

Answer 1

Is it supposed to be 'evaluate g(-1)'?

Answer 2

idk im not sure

Answer 3

hope this help

Answer 4

Yes, it does.

This just means to plug in 'a - 1' into the equation for 'x'.

\(g(x) = x^2 - x + 3\)

\(g(a - 1) = (a - 1)^2 - (a - 1) + 3\)

Can you simplify that?

Answer 5

First, can you simplify this?

\((a - 1)(a - 1)\)

Answer 6

a-1

Answer 7

No, we're multiplying them together..

\(a \times a = a^2\)

\(a \times -1 = -a\)

\(-1 \times a = -a\)

\(-1 \times -1 = 1\)

We multiply all of these to simplify it.

So we have:

\(a^2 - a - a + 1\)

Combine -a - a:

\(a^2 - 2a + 1\)

Plug this back into the function:

\(g(x) = a^2 - 2a + 1 - a - 1 + 3\)

What's 1 - 1? @cutegirl

Answer 8

0

Answer 9

Yes, and what's 0 + 3?

Answer 10

3

Answer 11

Yes, so we have:

\(g(a - 1) = a^2 - 2a - a + 3\)

Can you subtract -2a - a?

Answer 12

-2?

Answer 13

No, \(-2 - a\) is the same thing as \(-2 - 1\) but we put an 'a' next to it..so what do youget?

Answer 14

*you get?

Answer 15

2a^2?

Answer 16

No, what's -2 - 1?

Answer 17

-3

Answer 18

Yes, so \(-2 - a\) is \(-3a\).

So we have:

\(g(a - 1) = a^2 - 3a + 3\)

That's your answer.

Answer 19

I mean \(a^2 - 3a + 5\)..lol.

Answer 20

thanks lol