Question

can someone please help!!!! will medal ...An expression is shown below:

f(x) = –16x2 + 60x + 16

Part A: What are the x-intercepts of the graph of the f(x)? Show your work.

Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work.

Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph.

Answer 1

I dont want someone to just tell me the answer i want to learn how to do this type of problem

Answer 2

@jim_thompson5910 could you help when you help someone you explain it very well :)

Answer 3

Part A: To find the x-intercepts, set y=f(x) to 0 and solve the equation for x.
You could quadratic formula or factoring if it factorable or completing the square.

Answer 4

ok so it would be 0=-16x^2 +60x+16

Answer 5

yep

Answer 6

you could make those numbers easier on the eyes if you divide both side by 4
giving you 0=-4x^2+15x+4

Answer 7

this expression on the right hand side of the equation is factorable if you want to factor it

Answer 8

ok i have that write next in my notes and then im lost

Answer 9

ok well do you know the quadratic formula?

Answer 10

\[0=ax^2+bx+c \implies x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 11

no i dont know how to do that yet :(

Answer 12

ok I guess we will try factoring
if you have ax^2+bx+c
you can look for two factors that have product ac and sum b.
that means multiply to be ac and add up to b

Answer 13

you have -4x^2+15x+4
a=-4
b=15
c=4
ac=-16

So we want to find two numbers that multiply to be -16 and add up to be 15

Answer 14

can you think of two such numbers

Answer 15

pairs of numbers that multiply to be -16:
1(-16)
-1(16)
2(-8)
-2(8)
4(-4)

do any of these pairs of factors add up to be 15?

Answer 16

16 and -1

Answer 17

perfect

Answer 18

so 15=16-1

Answer 19

so we can write our middle term 15x
as 16x-1x
why does this help?

Answer 20

we have unlock the perfect sum such that we can factor by grouping

Answer 21

\[-4x^2+15x+4 \\ \\-4x^2+16x-1x+4 \\ (-4x^2+16x)+ -1x+4 \\ (-4x^2+16x)+(-1x+4)\]

Answer 22

now looking at the first group there
what is the gcf of the terms inside the first ( )

Answer 23

4

Answer 24

anything else be sure to look at the number of factors of x in each term

Answer 25

sorry 4x

Answer 26

cool stuff

Answer 27

so we will factor 4x from the first group

Answer 28

\[-4x^2+15x+4 \\ \\-4x^2+16x-1x+4 \\ (-4x^2+16x)+ -1x+4 \\ (-4x^2+16x)+(-1x+4) \\ 4x(-1x+4)+(-1x+4)\]

Answer 29

so 4x(-x+4)

Answer 30

you should now see two terms 4x(-1x+4) and (-1x+4)
and in these terms you should see the common factor (-1x+4)

Answer 31

wow your fast

Answer 32

\[-4x^2+15x+4 \\ \\-4x^2+16x-1x+4 \\ (-4x^2+16x)+ -1x+4 \\ (-4x^2+16x)+(-1x+4) \\ 4x(-1x+4)+(-1x+4) \\ 4x(-1x+4)+1(-1x+4)\]
I'm going to put a 1 next to our next group because people forget the 1 is there sometimes

Answer 33

now you are going to factor out the (-1x+4)
and yes you can it (-x+4) if you want :)

Answer 34

Rember ba+ca=(b+c)a or a(b+c)

example (kitty)(dog)+(kitty)(fish)=(kitty)(dog+fish)

Answer 35

so can show me what you think the factored expression looks like

Answer 36

(4x+1) (-x+4) (-x+4)

Answer 37

you have one too many factors

Answer 38

-x+4

Answer 39

need to be taken out

Answer 40

\[4x(-x+4)+1(-x+4) \\ \color{blue}{4x}\color{red}{(-x+4)} \color{blue}{+1 }\color{red}{(-x+4)} \\ \color{red}{(-x+4)}\color{blue}{(4x+1)}\]
right

Answer 41

so we did have an equation earlier
we were only playing with the right hand expression to get it in factored form

Answer 42

so we had
\[0=-16x^2+60x+16 \\ 0=-4x^2+15x+4 \\ 0=(-x+4)(4x+1)\]

Answer 43

now you set both factors equal to 0
because if you have something times something = 0 then at least one of those somethings is zero

Answer 44

so solve the equations:
-x+4=0 , 4x+1=0

Answer 45

so the first on i have x=4 or -x=-4 the second one i got x= -1/4

Answer 46

(x,y) is an ordered pair

remember we let y be 0 <-- this was to find the x-intercepts
So we know our answer should in the form (x,0)

Answer 47

So your answers for the first part is:
The x-intercepts are (4,0) and (-1/4,0)

Answer 48

Did you have any questions on the first part?

Answer 49

no you explained where i was having trouble

Answer 50

If you don't know the formula for the vertex...
Then you must be able to complete the square to find the vertex...
\[y=-16x^2+60x+16 \]


I'm going to assume we don't know the formula and just find the vertex form for this parabola which entails the completing the square thing I was talking about

\[y=-16x^2+60x+16 \\ y=(-16x^2+60x)+16\]
We are going to look at these first two terms only for now...
We are going to factor the number tagged onto the x^2 from both terms in the ( )

Answer 51

that is we are going to factor out -16 from both terms inside the ( )

Answer 52

you know -16/-16=1
\[y=(-16x^2+60x)+16 \\ y=\frac{-16}{-16}(-16x^2+60x)+16 \\ \\ y=-16(\frac{1}{-16})(-16x^2+60x)+16 \\ y=-16(\frac{-16x^2+60x}{-16})+16 \\ y=-16(\frac{-16}{-16}x^2+\frac{60}{-16}x)+16 \]
I wrote it like this on purpose because a lot of students have trouble seeing how to factor numbers out of other numbers where those numbers don't divide the other number evenly
that is for example in this question you do not see a factor of -16 in 60

Answer 53

anyways let's go ahead and reduce those fractions inside the ( )

Answer 54

\[y=-16(1x^2+\frac{-15}{4}x)+16 \\ y=-16(x^2+\frac{-15}{4}x)+16\]
let me know if you understand everything up til this point

Answer 55

ok

Answer 56

time to add something inside the grouping symbol so we can complete the square now

now whatever we add into that grouping symbol we are going to have remember that thing is being multiplied by the -16 on the outside
so whatever we put in we had to take out
I will show you what I mean in a bit

Answer 57

to complete the square
you need to have the coefficient of x^2 is 1
we have that now we just did that...
second we take the number in front of x....
divide by 2 and then square
and that is how we figure out what number to add inside the ( )

Answer 58

example
\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2\]

Answer 59

\[x^2+\frac{-15}{4}x+.... =(x+......)^2\]

Answer 60

so we take the number in front of x:
divide it by 2...(or you can say multiply by 1/2)
what do you get when you do this?

Answer 61

-15/8 i think

Answer 62

you think right

Answer 63

yeah

Answer 64

ok then square...
\[x^2+\frac{-15}{4}x+(\frac{-15}{8})^2 =(x+\frac{-15}{8})^2 \\ \text{ just like in the above formula I wrote } \\ x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \text{ note: where } k=\frac{-15}{4}\]

Answer 65

ok let's go back to our problem...
\[y=-16(1x^2+\frac{-15}{4}x)+16 \\ y=-16(x^2+\frac{-15}{4}x)+16 \\ y=-16(x^2+\frac{-15}{4}x+?)+16+16?\]
that ? we are fixing to plug in is (-15/8)^2
but before we do that I just want you to see that we are actually adding in 0...
notice that first ? is being multiplied by -16
so we are actually putting in -16? so we have to put in 16?
notice-16?+16? is =0

Answer 66

\[y=-16(x^2+\frac{-15}{4}x+(\frac{-15}{8})^2) +16+16(\frac{-15}{8})^2\]

Answer 67

now again the whole reason we did this so we could group the terms with x in it together
and then write it as a square

Answer 68

\[y=-16(x+\frac{-15}{8})^2+16+16(\frac{-15}{8})^2\]

Answer 69

you can add the like terms there
that is you can simplify:
\[16+16(\frac{-15}{8})^2\]

Answer 70

once you do that it should be really clear this is in vertex form
\[y=a(x-h)^2+k \text{ where vertex is } (h,k)\]
also if a is negative then (h,k) is a max
if a is positive then (h,k) is a min

Answer 71

I will give you time to digest before saying anything else

Answer 72

so then would you find out what (-15/8)^2 is

Answer 73

which is 225/64

Answer 74

yes
\[16+16(\frac{225}{64})=...\]

Answer 75

would i multiply 16 times 225 and 16 time 64 next

Answer 76

nope

Answer 77

a(b/c) is equal to ab/c
not ab/(ac)

Answer 78

\[16+\frac{16 \cdot 225}{64}\]

Answer 79

16+3600/64

Answer 80

and the end you should get 289/4 after adding those fractions and reducing

Answer 81

or you can reduce that one fraction then add

Answer 82

\[y=-16(x-\frac{15}{8})^2+\frac{289}{4}\]

Answer 83

so you should be able to identify the vertex

Answer 84

remember y=a(x-h)^2+k tells us the vertex is (h,k)
this also tells us if a is positive then (h,k) is min
and that if a is negative then (h,k) is max

Answer 85

please let me know what you think the vertex is

Answer 86

so because 16 is negative it is max

Answer 87

since -16=a
this means a is negative
so the vertex will be a max

Answer 88

so yes I would change what you said to
"so because -16 is negative it is max"

Answer 89

ok :}

Answer 90

Thank you for taking your time to help me on this problem im going to have to finish it up tomorrow but thank you so much for your help you were great !!!!!!!!

Answer 91

The cool thing about this problem is that all of the hard math is out of the way:
You are only left with the following things comparing y=-16(x-15/8)^2+289/4
to y=a(x-h)^2+k to identify (h,k) which is the vertex

And then graphing by plotting your x-intercepts and vertex and connecting the dots forming a parabola

Answer 92

Anyways
goodnight and good luck