Question

need help on the attach problem

Answer 1

ok you have a second degree polynomial so it looks like
\[p(x)=ax^2+bx+c\] and you know
\[p'(x)=2ac+b\] and
\[p''(x)=2a\]
now you are told that
\[p''(2)=3\] so
\[2a=3\] making
\[a=\frac{3}{2}\]

Answer 2

there is a typo above. it should be
\[p'(x)=2ax+b\]

Answer 3

so now we know
\[a=\frac{3}{2}\] and therefore
\[p'(x)=3x+b\]
and since
\[p'(2)=2\]we know
\[3\times 2+b=2\]
\[6+b=2\]
\[b=-4\]

Answer 4

Thanks

Answer 5

wait wait i was wrong on the last one

Answer 6

also why is there 2ac shouldn't be 2ax

Answer 7

we have
\[p(x)=\frac{3}{2}x^2-4x+c\] and
\[p(2)=6\] so
\[\frac{3}{2}\times 2^2-4\times 2+c=6\] not what i wrote earlier

Answer 8

that was a typo, sorry. i thought i wrote that. it is
\[p'(x)=2ax+b\]

Answer 9

oh, ok

Answer 10

i hit the "c" key instead of the "x" key because they are next to each other and my typing skillz are not what they should be

Answer 11

so is the value of P(1)=6?

Answer 12

Also plugging 2 into each x variable will give you 6, right?

Answer 13

oh i didn't complete it. what do you get for c?

Answer 14

is c = 8?

Answer 15

yes

Answer 16

However there no answers choices that match?

Answer 17

ok let me go back and check. it looks like
\[p(x)=\frac{3}{2}x^2-4x+8\] right?

Answer 18

well if that is right then
\[p(1)\]is a fraction for sure. lets check again

Answer 19

\[p(x)=ax^2+bx+c\]
\[p'(x)=2ax+b\]
\[p''(x)=2a\] and
\[p''(2)=3\] so
\[2a=3\] and
\[a=\frac{3}{2}\] all this is correct yes?

Answer 20

yes

Answer 21

\[p'(x)=2\times\frac{ 3}{2}x+b=3x+b\] and
\[p'(2)=3\times 2+b=2\] so
\[6+b=2\] and
\[b=-4\] this is ok right?

Answer 22

and finally
\[p(x)=\frac{3}{2}x^2-4x+c\] and
\[p(2)=\frac{3}{2}\times 2^2-4\times 2+c=6\]
\[6-8+c=6\]
\[c=-8\]

Answer 23

oh no
\[c=8\] but we had that before

Answer 24

so we now know that
\[p(x)=\frac{3}{2}x^2-4x+8\] and we are done. but you are right, p(1) is not in that list, but i am going to stick with this answer. i see no mistakes

Answer 25

where did the problem come from?

Answer 26

online website called quest, but where did 3x+b come from, and it should be one of the answer choices

Answer 27

Also suppose to find the value of P (1).

Answer 28

right and i don't see p(1) in your list. the second derivative of a quadratic is a constant. it is
\[p''(x)=2a\]

Answer 29

since you are told that
\[p''(2)=3\] that makes
\[2a=3\] so
\[a=\frac{3}{2}\]

Answer 30

the first derivative is
\[2ax+b\] and since we know
\[a=\frac{3}{2}\]we know
\[2a=3\]so the first derivative is
\[p'(x)=3x+b\]

Answer 31

i reposted this problem so lets see if we get a different answer

Answer 32

ok