Question

the axis of symmetry for the graph of the function is f(x) = 1/4x^2 + bx + 10 is x = 6. What is the value of b?

Answer 1

axis-of symmetry := the vertical line about which the parabola is symmetric
(the vertical line that splits the parabola into 2 equal parts)

You need to complete the square (in terms of b), and find the x-coordinate of the vertex (in terms of b), and set this x-coordinate of the vertex equal to 6, to solve for b.

Answer 2

And vertex can be found through «completing the square»

Answer 3

u want me to plug in the x first?

Answer 4

I want you to complete the square.

Answer 5

1/16?

Answer 6

EXAMPLE\( : \)

If the axis of symmetry for the
function \(\color{#000000 }{ \displaystyle f(x)=2x^2+bx+8 }\)
is, \(\color{#000000 }{ \displaystyle x=4 }\), then find \(\color{#000000 }{ \displaystyle b }\).


SIMPLIFICATION\( : \)
The \(\color{#000000 }{ \displaystyle +8 }\) in the \(\color{#000000 }{ \displaystyle f(x) }\) is just a vertical shift (up),
therefore, you can tell that the axis of symmetry
is going to be the same for \(\color{#000000 }{ \displaystyle f(x)=2x^2+bx+8 }\)
and for \(\color{#000000 }{ \displaystyle f(x)=2x^2+bx }\).

So, we can just complete the square by,
\(\color{#000000 }{ \displaystyle f(x)=2x^2+bx }\)


QUESTION\( : \)
\(\color{#000000 }{ \displaystyle f(x)=2\left(x^2+\frac{b}{2}x\right) }\) (factored out of 2)
Now, you need to find the number that will make
your expression inside the parenthesis into a
perfect square trinomial (when added into parenthesis).


RULE/CONCEPT\( : \)
In general, \(\color{#000000 }{ \displaystyle x^2+2ax+a^2=(x+a)^2 }\)
is a perfect square trinomial. After comparing, you can
tell that in your case, \(\color{#000000 }{ \displaystyle 2a=\frac{b}{2} }\), and therefore,
\(\color{#000000 }{ \displaystyle a=\frac{b}{4} \quad \Rightarrow \quad a^2=\frac{b^2}{16} }\)


COMPLETING THE SQUARE\( : \)

\(\color{#000000 }{ \displaystyle f(x)=2\left(x^2+\frac{b}{2}x\right) }\)

\(\color{#000000 }{ \displaystyle f(x)=2\left(x^2+\frac{b}{2}x+\frac{b^2}{16}-\frac{b^2}{16}\right) }\)

\(\color{#000000 }{ \displaystyle f(x)=2\left(x^2+\frac{b}{2}x+\frac{b^2}{16}\right)-2\times \frac{b^2}{16} }\)

\(\color{#000000 }{ \displaystyle f(x)=2\left(x^2+\frac{b}{2}x+\frac{b^2}{16}\right)-2\times \frac{b^2}{16} }\)

Recall that you really have;

\(\color{#000000 }{ \displaystyle f(x)=2\left(x^2+2\left(\frac{b}{4}\right)x+\left(\frac{b}{4}\right)^2\right)-2\times \frac{b^2}{16} }\)

in a form of general perfect square trinomial, with a=b/4;
So you end up with
\(\color{#000000 }{ \displaystyle f(x)=2\left(x+\frac{b}{4}\right)^2-2\times \frac{b^2}{16} }\)

So your x-axis (in terms of b) is:
\(\color{#000000 }{ \displaystyle x{\rm~of~symm:}\quad \quad h=b/4 }\)
But, you are also given that the axis of symmetry is,
\(\color{#000000 }{ \displaystyle x{\rm~of~symm:}\quad \quad h=4 }\)

So,
\(\color{#000000 }{ \displaystyle h=b/4,\quad h=4 \quad\quad \Rightarrow \quad4=b/4\quad \Rightarrow \quad \color{blue}{16=b}}\)

Answer 7

This can certainly be written shorter; this isn't a lot, don't be afraid...
(I just elaborated everything)

Answer 8

O_O ok i'll... i'll try

Answer 9

{-_=]

Answer 10

ohhhhhhh

Answer 11

thanks

Answer 12

?

Answer 13

https://www.youtube.com/watch?v=WpVChAg2hRg

Answer 14

this i me drag racing

Answer 15

im the red one

Answer 16

dude, how old are u

Answer 17

12

Answer 18

ah ok

Answer 19

ಠ_ಠ

Answer 20

{0___0]

Answer 21

closing this in 5

Answer 22

4

Answer 23

3

Answer 24

2

Answer 25

ilots that rac the first time then i came back and bett her

Answer 26

1

Answer 27

oh ok