Question

Three functions are given below: f(x), g(x), and h(x). Explain how to find the axis of symmetry for each function, and rank the functions based on their axis of symmetry (from smallest to largest).
f(x) = –4(x − 8)2 + 3
g(x) = 3x2 + 12x + 15

Answer 1

This

Answer 2

@pooja195

Answer 3

@Nnesha

Answer 4

\[\huge\rm y=a(x-\color{red}{h})^2+\color{blue}{k}\] vertex form of quadratic equation
where (h,k) is the vertex and x-coordinate of the vertex represent x-axis of symmetry
axis of symm. = h

Answer 5

ok.. so how do you find it like plug in this formula?

Answer 6

if the equation is already in vertex form then you don't need to plugin anything
just look at the equation what number is replaced by h ?

if the given equation is standard then u can use this formula to find x-coordinate of the vertex \[\large\rm x=-\frac{b}{2a}\]
b= coefficient of x term
a=leading coefficient
\[\rm y=ax^2+bx+c\]

Answer 7

standard form*

Answer 8

f(x) = –4(x − 8)2 + 3 8 is replaced by h

Answer 9

so axis of symmetry for f(x) would be 8?

Answer 10

yes right.

Answer 11

for g(x) do you have to plug in the formula?

Answer 12

yes right

Answer 13

you just rreplace the numbers or you have to solve it or something?

Answer 14

first replace the number with their values and then simplify

Answer 15

would it be g(x)= 3(x-12)^2 +15?

Answer 16

hmm how did convert g(x) standard form to vertex ?
are you familiar with the `completing the square ` method ?

Answer 17

yaa

Answer 18

the formula i gave you is simple and easier than completing the square
it will take just few minutes
but it's okay if you want to complete the square

Answer 19

\(\color{blue}{\text{Originally Posted by}}\) @Janu16
would it be g(x)= 3(x-12)^2 +15?
\(\color{blue}{\text{End of Quote}}\)
that's not correct show me the work how did you get that equation :=))

Answer 20

ohh i just replaced the numbers. I didn't simplify

Answer 21

no you can't just replace number
to arrive from \[\rm Ax^2+Bx+C \] to \[\rm a(x-h)^2+k\] you have to apply completing the square method you can just replace the numbers around

Answer 22

use this formula to find x-coordinate of vertex (which is equal to axis of symm)\[\rm x=-\frac{b}{2a}\] replace b and a in this formula by their values

Answer 23

i don't know how to use that formula. if you don't mind can you help solve this with completing the sqaure cause thats what im learning

Answer 24

ohh sure!
for that formula we just have to replace b with 12 and a with 3
\[\large\rm y=\color{reD}{a}x^2+\color{blue}{b}x+\color{green}{c}\]\[\large\rm y=\color{reD}{3}x^2+\color{blue}{12}x+\color{green}{15}\]
a=leading coefficient
b=middle term( coefficient of x term)
c=constant term

Answer 25

let's complete the square
\[\large\rm g(x)=3x^2+12x+15\]
since the leading coefficient isn't one it would be great if we first take out the GCF(greatest common factor ) of first two terms

Answer 26

\[\rm g(x) = \color{Red}{(3x^2+12x)}+15\]
we aren't gonna touch the constant term till the end

Answer 27

what's the greatest common factor in first two terms ?

Answer 28

4

Answer 29

wait no

Answer 30

3

Answer 31

right take out 3 from (3x^2 +12x) what would u have left with ?

Answer 32

x+4

Answer 33

hmm just take out the common factor 3 not the x variable
the reason is the to get the 1 for leading coefficient so it would be \[\large\rm g(x) =3(\color{reD}{x^2+4x})+15\]now we are going to complete the square of `x^2+4x`
1st) we should take `half of x term` and then square it

Answer 34

half of 4 would be 2

Answer 35

\[\huge\rm y=(\color{ReD}{x^2+bx})+c\]
i would take of of x term which is b in this example
and then subtract of (b/2)^2 from constant term
\[\large\rm y=(\color{reD}{x^2+bx} +(\frac{b}{2})^2)+c - (\frac{b}{2})^2\]

Answer 36

remember the factors of this quadratic equation would always be \[\rm (x+\frac{b}{2})^2\]
(x+ half of b term)^2

Answer 37

\[\large\rm y=(\color{blue}{x^2+bx +(\frac{b}{2})^2})+c - (\frac{b}{2})^2\]
\[\large\rm y=\color{blue}{(x+\frac{b}{2})^2} +c -(\frac{b}{2})^2\]

Answer 38

so\[(x^2+4x)15-4\]

Answer 39

is that correct^^

Answer 40

or is it (x2+4x + 4)15−4

Answer 41

well there is something missing
you should add (b/2)^2 in the parentheses to make it trinomial equation

Answer 42

that looks good but don't forget the sign +15-4

Answer 43

hmm but wait we have common factor at front og the parentheses

Answer 44

\[\large\rm g(x) =3(\color{reD}{x^2+4x+4})+15\]
when we have a GCF we should multiply (b/2)^2 by gcf and THEN subtract from constant term

Answer 45

so 4 times 4?

Answer 46

no (b/2)^2 = ??

Answer 47

4

Answer 48

8

Answer 49

which one ?? :=))
b is what ?

Answer 50

so you do 4 divided by 2 which is 2 than sqaure it which is 4

Answer 51

yes right 4 is correct 8 isn't now multiply 4 times 3(gcf) then subtract from the constant term

Answer 52

so you 4 times 3 which is 12

Answer 53

than 15-12?

Answer 54

right

Answer 55

so what would be the final equation ?

Answer 56

\[x^2+4x=3\]

Answer 57

hmm no

Answer 58

we dont write the 3 in the beginning because we got rid of it right?

Answer 59

we didn't get rid of it just multiply it by (b/2)^2 to subtract it from constant term

Answer 60

x^2+4x+3=0

Answer 61

not really we are working on one side so \[\large\rm g(x) =3(\color{reD}{x^2+4x+4})+15\color{blue}{-12}\]
remember we got the trinomial equation

Answer 62

ok..

Answer 63

and the factor of that trinomial would always be \[\rm (x+\frac{b}{2})^2\]

Answer 64

so what would the equation look like if you multiply 4 and 3 and subtract 15 from 12?

Answer 65

15-12= 3
right now deal with this equation (x^2+4x+4) what are the factors ?

Answer 66

like i said when we complete teh square the factors of the trinomial would always be \[\rm (x+\frac{b}{2})^2\]
b= coefficient of x term

Answer 67

does it make sense ?

Answer 68

yaa

Answer 69

alright then what would be the factor of (x^2+4x+4) ?

Answer 70

lie factoring the whole thing?

Answer 71

x+2^2

Answer 72

factor the trinomial (quadratic equation )(red part in this equatin) \[\large\rm g(x) =3(\color{reD}{x^2+4x+4})+15\color{blue}{-12}\]
( quadratic equation )

Answer 73

not just 2 to the 2 power

Answer 74

remember u will always get \[\rm (x+\frac{b}{2})^2\] in that equation b is 4 so \[(x+\frac{4}{2})^2 \rightarrow (x+2)^2\]

Answer 75

so what would be the final equation ?

Answer 76

for h(x) remember