Question

Please help me easy algebra
f(x)= 4x^2-x+4

Find 2f(a)
and
[f(a)]^2

Answer 1

@jim_thompson5910

Answer 2

You're a bus +_+

Answer 3

I like AmTran buses, I am one!

Answer 4

\[\Large\rm f(\color{royalblue}{x})=4\color{royalblue}{x}^2-\color{royalblue}{x}+4\]If we multiply by 2,\[\Large\rm 2f(\color{royalblue}{x})=2(4\color{royalblue}{x}^2-\color{royalblue}{x}+4)\]And then plug a in for our x,\[\Large\rm 2f(\color{royalblue}{a})=2(4\color{royalblue}{a}^2-\color{royalblue}{a}+4)\]

Answer 5

Thank you! Makes sense!

Answer 6

So then uhhhh, how bout the next one?
Shouldn't be too bad, right?

Plug in a, then square each side.

Answer 7

The [()] messes me up.

Answer 8

@zepdrix

Answer 9

They just added another set of brackets because \(\Large\rm f(a)^2\) looks really sloppy.

Answer 10

There is no difference between square and round brackets.
We just use different brackets to make it easier to match them up sometimes.

Answer 11

Thank you.

Answer 12

\[\Large\rm f(\color{royalblue}{x})=4\color{royalblue}{x}^2-\color{royalblue}{x}+4\]So plugging in a gives you,\[\Large\rm f(\color{royalblue}{a})=4\color{royalblue}{a}^2-\color{royalblue}{a}+4\]Squaring each side gives you,\[\Large\rm \left(f(\color{royalblue}{a})\right)^2=\left(4\color{royalblue}{a}^2-\color{royalblue}{a}+4\right)^2\]See how the double round brackets are a little harder to read?

Answer 13

From there, your teacher might want you to expand out the brackets.. I'm not sure :o

Answer 14

Can you help me find this domain?

Find domain of y=(1/2x)-(11/2)

Answer 15

Find domain of y=(1/2x)-(11/2)

Answer 16

Can you help me with that?

Thanks for working that other out.