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

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DanJS
 one year ago
Best ResponseYou've already chosen the best response.0do you se the general shape of it by the form , it is already in the general form

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0f(x) = ax^2 + bx + c general quadratic form, parabola

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do i find the domain for the graph ? i factorized it and i got (x+2)^2 + 9

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0watch the completing the square there, you did that part right, but the constant 9 is wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it 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
 one year ago
Best ResponseYou've already chosen the best response.0given m(x) = x^2 + 4x + 9 Graph the function and find the domain? what is it asking

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its asking the value of a

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh and i got the completing square... (x+2)^2 + 5

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0the domain of that is going to be all real numbers for possible values of X

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0It states that x is a real number because you want to exclude complex numbers or roots

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but how do i draw the graph...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where will the intersection for the parabola be

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oki i finally got it!!

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0I would first find the vertex point, the x coordinate of that will be found by calculating b / (2*a) = 4/2(1) = 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry i got confused in the vertex point part..

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0after 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 xaxis intercepts

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0if it was x^2 it woul dopen down, and you would have two xintercepts to find

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it always found using b / 2*a??

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0you 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
 one year ago
Best ResponseYou've already chosen the best response.0oh i didn't know that before... isn't it usually b +/ (sqrt.) (b^2 4ac) / 2a

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0that is the quadratic formula, used in finding the values where the graph crosses the xaxis, or the roots of the quadratic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oki so to find the vertex i always use the formula used for x^2 graphs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats b/ 2a as u stated before

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0notice the stuff under the root term turns out to be a negative number b^24ac no roots in the real numbers, cant have a square root of a negative number

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0that is why you don't see it cross the x axis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think i get it now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0once again thanks for the help :)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0to graph a parabola, i would find the easy points first vertex xintercepts if any yintercepts if any that is probably enough to get the general graph

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0if 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
 one year ago
Best ResponseYou've already chosen the best response.0I think that is what they were talking about the x >a thing, but this one does not open sideways, so no 'a' value

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0The domain of that will be all x values greater than or equal to zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is there a reflection to this graph? m ^1 (x)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0yes, that would be the graph resulting in reflecting over the line y = x

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0x = y^2 + 4y + 9 you can resolve for y if you want, but that is the inverse

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i know how to resolve it.... but the graph drawing part is annoying

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0you are basically switching the x and y values in every point

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0the new vertex is (5 , 2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0u only switch the vertex?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about the coordinates (0, 9)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0the + 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
 one year ago
Best ResponseYou've already chosen the best response.0ok, 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
 one year ago
Best ResponseYou've already chosen the best response.0so all reflections are either to the right or left for an parabolic graph?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its fine i think i got it... thank u so much :)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0the inverse function is a reflection over both the x and the y axis, combined it is a reflection over the line y = x

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0swap the variables x and y in the function and then solve for y, that will be the inverse , if it exists

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0every 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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