anonymous
  • anonymous
how do i draw the graph for m (x)= x^2 + 4x + 9
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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DanJS
  • DanJS
do you se the general shape of it by the form , it is already in the general form
DanJS
  • DanJS
f(x) = ax^2 + bx + c general quadratic form, parabola
anonymous
  • 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
  • DanJS
watch the completing the square there, you did that part right, but the constant 9 is wrong
anonymous
  • 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
  • DanJS
given m(x) = x^2 + 4x + 9 Graph the function and find the domain? what is it asking
anonymous
  • anonymous
its asking the value of a
anonymous
  • anonymous
in which x > a
anonymous
  • anonymous
oh and i got the completing square... (x+2)^2 + 5
DanJS
  • DanJS
the domain of that is going to be all real numbers for possible values of X
DanJS
  • DanJS
It states that x is a real number because you want to exclude complex numbers or roots
anonymous
  • anonymous
but how do i draw the graph...
anonymous
  • anonymous
where will the intersection for the parabola be
DanJS
  • DanJS
|dw:1443165464570:dw|
anonymous
  • anonymous
oki i finally got it!!
anonymous
  • anonymous
thank you so much
DanJS
  • 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
  • DanJS
hah, you sure,, kk
anonymous
  • anonymous
I'm sorry i got confused in the vertex point part..
DanJS
  • 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
  • DanJS
if it was -x^2 it woul dopen down, and you would have two x-intercepts to find
anonymous
  • anonymous
is it always found using -b / 2*a??
DanJS
  • 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
  • DanJS
x = -b/(2a)
DanJS
  • 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
  • anonymous
oh i didn't know that before... isn't it usually -b +/- (sqrt.) (b^2 -4ac) / 2a
DanJS
  • 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
  • anonymous
oki so to find the vertex i always use the formula used for x^2 graphs
anonymous
  • anonymous
thats -b/ 2a as u stated before
DanJS
  • 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
  • anonymous
yep so its ignored?
DanJS
  • DanJS
that is why you don't see it cross the x axis
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
i think i get it now
anonymous
  • anonymous
once again thanks for the help :)
DanJS
  • 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
  • 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
  • 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
  • DanJS
\[x = y^2\]
DanJS
  • DanJS
The domain of that will be all x values greater than or equal to zero
anonymous
  • anonymous
is there a reflection to this graph? m ^-1 (x)
DanJS
  • DanJS
yes, that would be the graph resulting in reflecting over the line y = x
DanJS
  • DanJS
x = y^2 + 4y + 9 you can re-solve for y if you want, but that is the inverse
anonymous
  • anonymous
yeah i know how to resolve it.... but the graph drawing part is annoying
DanJS
  • DanJS
|dw:1443166512627:dw|
DanJS
  • DanJS
you are basically switching the x and y values in every point
DanJS
  • DanJS
the new vertex is (5 , -2)
anonymous
  • anonymous
u only switch the vertex?
anonymous
  • anonymous
what about the coordinates (0, 9)
DanJS
  • DanJS
|dw:1443166710633:dw|
DanJS
  • 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
  • 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
  • anonymous
so all reflections are either to the right or left for an parabolic graph?
anonymous
  • anonymous
its fine i think i got it... thank u so much :)
DanJS
  • 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
  • DanJS
swap the variables x and y in the function and then solve for y, that will be the inverse , if it exists
DanJS
  • DanJS
|dw:1443167505039:dw|
DanJS
  • 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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