how do i draw the graph for m (x)= x^2 + 4x + 9

- anonymous

how do i draw the graph for m (x)= x^2 + 4x + 9

- schrodinger

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- DanJS

do you se the general shape of it by the form , it is already in the general form

- DanJS

f(x) = ax^2 + bx + c
general quadratic form, parabola

- anonymous

how do i find the domain for the graph ?
i factorized it and i got (x+2)^2 + 9

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- DanJS

watch the completing the square there, you did that part right, but the constant 9 is wrong

- anonymous

it says that x is a real number... obviously cuz the graph is infinity
but it also says that x> a and i have to find the value of a.... how do i find a?

- DanJS

given m(x) = x^2 + 4x + 9
Graph the function and find the domain? what is it asking

- anonymous

its asking the value of a

- anonymous

in which x > a

- anonymous

oh and i got the completing square... (x+2)^2 + 5

- DanJS

the domain of that is going to be all real numbers for possible values of X

- DanJS

It states that x is a real number because you want to exclude complex numbers or roots

- anonymous

but how do i draw the graph...

- anonymous

where will the intersection for the parabola be

- DanJS

|dw:1443165464570:dw|

- anonymous

oki i finally got it!!

- anonymous

thank you so much

- DanJS

I would first find the vertex point, the x coordinate of that will be found by calculating
-b / (2*a) = -4/2(1) = -2

- DanJS

hah, you sure,, kk

- anonymous

I'm sorry i got confused in the vertex point part..

- DanJS

after you find the vertex, notice the constant on the x^2 term is positive, meaning it opens upwards, and starting at that vertex, there will be no x-axis intercepts

- DanJS

if it was -x^2 it woul dopen down, and you would have two x-intercepts to find

- anonymous

is it always found using -b / 2*a??

- DanJS

yes, the x coordinate of the vertex, then use that value in the function to get the y coordinate of the point.
Do you need to see how that formula was found or no
?

- DanJS

x = -b/(2a)

- DanJS

you were trying to put the function into the 'vertex form' by completing the square. that is possible too, and you can read the point right from the equation

- anonymous

oh i didn't know that before... isn't it usually -b +/- (sqrt.) (b^2 -4ac) / 2a

- DanJS

that is the quadratic formula, used in finding the values where the graph crosses the x-axis, or the roots of the quadratic

- anonymous

oki so to find the vertex i always use the formula used for x^2 graphs

- anonymous

thats -b/ 2a as u stated before

- DanJS

notice the stuff under the root term turns out to be a negative number b^2-4ac
no roots in the real numbers, cant have a square root of a negative number

- anonymous

yep so its ignored?

- DanJS

that is why you don't see it cross the x axis

- anonymous

oh ok

- anonymous

i think i get it now

- anonymous

once again thanks for the help :)

- DanJS

to graph a parabola, i would find the easy points first
vertex
x-intercepts if any
y-intercepts if any
that is probably enough to get the general graph

- DanJS

if it is y in terms of x, and opens up or down, the domain is all real numbers
if it is x in terms of y, and opens right or left, then the domain will be all x to the right or left of the value x=a, the vertex point x value

- DanJS

I think that is what they were talking about the x >a thing, but this one does not open sideways, so no 'a' value

- DanJS

\[x = y^2\]

- DanJS

The domain of that will be all x values greater than or equal to zero

- anonymous

is there a reflection to this graph? m ^-1 (x)

- DanJS

yes, that would be the graph resulting in reflecting over the line y = x

- DanJS

x = y^2 + 4y + 9
you can re-solve for y if you want, but that is the inverse

- anonymous

yeah i know how to resolve it.... but the graph drawing part is annoying

- DanJS

|dw:1443166512627:dw|

- DanJS

you are basically switching the x and y values in every point

- DanJS

the new vertex is (5 , -2)

- anonymous

u only switch the vertex?

- anonymous

what about the coordinates (0, 9)

- DanJS

|dw:1443166710633:dw|

- DanJS

the + value on the y^2 term means it opens to the right
the domain in that case for the inverse is all values of x such that x is larger or equal to 5

- anonymous

ok, so for refection we make the subject of the formula x so the other side has the y term... and as the y is positive we open it to the right side... if it was negative it would be the left?

- anonymous

so all reflections are either to the right or left for an parabolic graph?

- anonymous

its fine i think i got it... thank u so much :)

- DanJS

the inverse function is a reflection over both the x and the y axis, combined it is a reflection over the line y = x

- DanJS

swap the variables x and y in the function and then solve for y, that will be the inverse , if it exists

- DanJS

|dw:1443167505039:dw|

- DanJS

every point on the upwards one has a corresponding point on the sideways inverse, the values are just the switched numbers for x and y

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