Question

PLEASE HELP!!!
find the vertex, axis of symmetry, the intercepts, the domain, the range, intervals where the function increases, intervals where the function decreases and graph y=x^2+4x

Answer 1

where would you like to start?

Answer 2

the vertex please

Answer 3

ok
the first coordinate of the vertex of \(y=ax^2+bx+c\) is \(-\frac{b}{2a}\)
in your case \(a=1,b=4\) so what is the first coordinate of the vertex?

Answer 4

or -2

Answer 5

not quite
right the second time

Answer 6

first coordinate of the vertex is \(-2\) and the second coordinate of the vertex is what you get for \(y\) when you replace \(x\) by \(-2\)

Answer 7

btw finding the vertex answer almost every other question above

Answer 8

I got a(-2)^2+4(-2)+0

Answer 9

-2a-8+0 ???

Answer 10

there is no \(a\) in this , it is just \((-2)^2+4(-2)\)

Answer 11

which give you \(4-8=-4\) for the \(y\) coordinate

Answer 12

ok no a because a=1?

Answer 13

yeah i just wrote \(y=ax^2+bx+c\) as a general form
your specific one is \(y=x^2+4x\)

Answer 14

ok
my vertex is (-2,-4)

Answer 15

right
now we can breeze through most of the other questions

Answer 16

axis of symmetry
since the first coordinate of the vertex is \(-2\) the axis of symmetry is \(x=-2\)

Answer 17

it's that simple?

Answer 18

the domain
since this is a polynomial, the domain is all real numbers

Answer 19

is the range y

Answer 20

yeah that simple
the range
since the second coordinate of the vertex is \(-4\) and this parabola opens up
the range is \([-4,\infty)\) or \(y\geq -4\)

Answer 21

ok did we skip the intercepts?

Answer 22

intervals where the function decreases
since the axis of symmetry is \(x=-2\) it decreases on \((-\infty, -2)\) or \(x<-2\)

Answer 23

so evidently increases on \(x>-2\)
vertex tells all and it is easy
skipped intercepts because that is not from the vertex, we have to do something else

Answer 24

that is why i skipped the intercepts
but they are not hard, it is just that they do not come from the vertex

Answer 25

ok so that's why I got it wrong...please explain

Answer 26

\(y\) intercept is easiest
that comes by putting \(x=0\) and you get if \(x=0\) then \(y=0\) and so the \(y\) intercept is 0

Answer 27

\(x\) intercepts put \(y=0\) and get
\[x^2+4x=0\] then solve for \(x\) which is not hard in this example

Answer 28

factor as \(x(x+4)=0\)and you see that the zeros are \(x=0\) or \(x=-4\)

Answer 29

you lost me

Answer 30

o I got it...so both 0nd -4 are the answers

Answer 31

lets go slow
we are finding the \(x\) intercepts, where the graph crosses the \(x\) axis

Answer 32

it crosses the \(x\) axis where \(y=0\) so we set \(x^2+4x=0\) and get \(x(x+4)=0\) making \(x=-4\) or \(x=0\)
the \(x\) intercepts, as ordered pairs, are \((-4,0)\) and \((0,0)\)

Answer 33

and the \(y\) intercept is also \((0,0)\)

Answer 34

now we got it all, right?

Answer 35

yes, so my graph will have 3 points

Answer 36

it is a parabola that opens up
the minimum point on the parabola is the vertex, which is at \((-2,-4)\) and it is symmetric about \(x=-2\) and crosses the \(x\) axis at \((-4,0)\) and \((0,0)\) aka the origin

Answer 37

this is the part that confuses me

Answer 38

thank you for your help

Answer 39

here is a nice picture

Answer 40

http://www.wolframalpha.com/input/?i=x^2%2B4x

Answer 41

you can see all the feature, decreasing, increasing, vertex, x intercepts etc

Answer 42

yw

Answer 43

thanks for the photo this really helps

Answer 44

good, wolfram is nice to use