Evaluating Function

Danoano

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Aug 23, 2019
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Hi all,

I hope this is the right place for this post.
I am trying to learn evaluating functions but I am stuck.

Given f (x) = 2x2 + 4x - 3, find f (2a + 3).

f (2a + 3) = 2(2a + 3)2 + 4(2a + 3) - 3
= 2(4a2 + 12a + 9) + 8a + 12 - 3
= 8a2 + 24a + 18 + 8a + 12 - 3
= 8a2 + 32a + 27

My problem comes at second row.
Could someone explain how do you get = 2(4a2 + 12a + 9) where does 12a come from ?
For me 2(2a + 3)2 = 2(4a+9) = 8a + 18
or when I multiply the brackets first before power I get 2(2a + 3)2 = (4a + 6)2 = 16a +36
What am I missing ?

Thank you
 
Last edited:
Hi all,

I hope this is the right place for this post.
I am trying to learn evaluating functions but I am stuck.

Given f (x) = 2x2 + 4x - 3, find f (2a + 3).

f (2a + 3) = 2(2a + 3)2 + 4(2a + 3) - 3
= 2(4a2 + 12a + 9) + 8a + 12 - 3
= 8a2 + 24a + 18 + 8a + 12 - 3
= 8a2 + 32a + 27

My problem comes at second row.
Could someone explain how do you get = 2(4a2 + 12a + 9) where does 12a come from ?
For me 2(2a + 3)2 = 2(4a+9) = 8a + 18
or when I multiply the brackets first before power I get 2(2a + 3)2 = (4a + 6)2 = 16a +36
What am I missing ?

Thank you
I believe your function is:

f(x) = 2x^2 + 4x -3

In computers (almost everywhere now - '^' indicates 'power' or 'exponent')

do you now the following identity:

(a + b)^2 = a^2 + 2ab + b^2
 
Hi all,

I hope this is the right place for this post.
I am trying to learn evaluating functions but I am stuck.

Given f (x) = 2x2 + 4x - 3, find f (2a + 3).

f (2a + 3) = 2(2a + 3)2 + 4(2a + 3) - 3
= 2(4a2 + 12a + 9) + 8a + 12 - 3
= 8a2 + 24a + 18 + 8a + 12 - 3
= 8a2 + 32a + 27

My problem comes at second row.
Could someone explain how do you get = 2(4a2 + 12a + 9) where does 12a come from ?
For me 2(2a + 3)2 = 2(4a+9) = 8a + 18
or when I multiply the brackets first before power I get 2(2a + 3)2 = (4a + 6)2 = 16a +36
What am I missing ?

Thank you
[MATH]f(x) = 2x^2 + 4x - 3 \implies[/MATH]
[MATH]f(2a + 3) = 2(2a + 3)^2 + 4(2a + 3) - 3.[/MATH]
What happens when you square (2a + 3).
 
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