anonymous
  • anonymous
Which of these shows the graph for the quadratic function y = x^2 + 2x + 3?
Algebra
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anonymous
  • anonymous
Which of these shows the graph for the quadratic function y = x^2 + 2x + 3?
Algebra
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
anonymous
  • anonymous

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anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
UsukiDoll
  • UsukiDoll
ok first of all do not spam and mass tag. I wouldn't want you to end up getting a warning or suspended as this is against OpenStudy's Code of Conduct and Terms of Service.
anonymous
  • anonymous
hey your picture is rolanda!
UsukiDoll
  • UsukiDoll
Everyone, I got this question under control, so anyway... we need to create a chart that looks like x y = x^2+2x+3 then we just select any x... and solve for that y.... once we have done that, we can pick more x's that will give more y = results and we will have more coordinate points.. For example, if my x = 0 Then x y = x^2+2x+3 0 y=0^2+2(0) +3 0 y =3 so or first coordinate point is (0,3). You graph that on your scratch paper... then you let x be any positive or negative number and continue until your graph matches one of the choices you are given.
anonymous
  • anonymous
no please stay i really need help
UsukiDoll
  • UsukiDoll
but there are mods in here and I'm not going to give direct answers... my task as a tutor is to teach and guide, not spoonfeed.
anonymous
  • anonymous
they are here to help too
UsukiDoll
  • UsukiDoll
I already gave you an example on how to solve for this ... please try to understand and attempt the work... then you can answer here and I'll check to see if you're on the right track or not.
anonymous
  • anonymous
i dont understand @UsukiDoll explanation is it possible for someone else to explain it like @ganeshie8 @Michele_Laino
UsukiDoll
  • UsukiDoll
@ganeshie8 @Michele_Laino... did I not just give an example and explain this thoroughly?
anonymous
  • anonymous
u did but im just not understanding it
Michele_Laino
  • Michele_Laino
I think you gave a good explanation @UsukiDoll :)
UsukiDoll
  • UsukiDoll
see....I'm being thrown under the bus here.
anonymous
  • anonymous
im not throwing u under the bus... i have a disease where i need everything to be explained extra cause i cant remember concepts and my brain is messed up cause it hurts to think
anonymous
  • anonymous
its a birth defect... :(
anonymous
  • anonymous
its called Cerebellar Ataxia
anonymous
  • anonymous
Michele_Laino
  • Michele_Laino
another way is: the x-ccordinate of the vertex of a parabola like this: \[y = a{x^2} + bx + c\] is: \[x = - \frac{b}{{2a}}\] now, by comparison with your parabola, what are the coefficient a, b, and c, and what is the x-coordinate of the vertex of your parabola?
anonymous
  • anonymous
whats a parabloa
UsukiDoll
  • UsukiDoll
there is more than one way to solve this...maybe that method is too complex for OP?
anonymous
  • anonymous
pleae help me:(
UsukiDoll
  • UsukiDoll
alright.. I'll try again... what she gave was from College Algebra and seeing as you're not at level yet, we'll go back to what I've just put earlier.
UsukiDoll
  • UsukiDoll
we are given \[y = x^2 + 2x + 3\] and the question asks which graph is the same as the quadratic equation
Michele_Laino
  • Michele_Laino
I'm sorry, please ignore my hint @barbdewysveck1
anonymous
  • anonymous
please help me :(
UsukiDoll
  • UsukiDoll
what we can do....is to create a table a table like x y =x^2+2x+3
UsukiDoll
  • UsukiDoll
so, we can pick any x and plug it in the y=x^2+2x+3
UsukiDoll
  • UsukiDoll
For example, if I pick x = 0. I replace all x's in the y equation x y=x^2+2x+3 0 y=(0)^2+2(0)+3 0^2, also means 0 written twice (0)(0) = 0 2(0) means 2 x 0 which is also 0 so now we have 0 0+0+3 (we add this together and that's 3) so you're first coordinate point is (0,3)
anonymous
  • anonymous
so b is the answer
UsukiDoll
  • UsukiDoll
well... we only have one point... we need more than (0,3) for a clearer picture
UsukiDoll
  • UsukiDoll
so we continue by picking another x... how about if x = 1 so I plug x = 1 into y=x^2+2x+3 1 y=(1)^2+2(1)+3 (1)^2 means (1)(1) = 1 x 1 = 1 2(1) means 2 x 1 and that's 2 so we add up 1+2+3 and we have y = 6 so our next point is (1,6)
UsukiDoll
  • UsukiDoll
|dw:1434428975864:dw|
UsukiDoll
  • UsukiDoll
I connect those two points after I graph them |dw:1434429072437:dw|
anonymous
  • anonymous
its A!!!!!!!!!
UsukiDoll
  • UsukiDoll
yes it is!
anonymous
  • anonymous
yay can u help me with more
UsukiDoll
  • UsukiDoll
because we started at (0,3) and by plotting (1,6) we noticed that the graph was really high up... so only 1 graph was the same as our drawing
UsukiDoll
  • UsukiDoll
sure
anonymous
  • anonymous
UsukiDoll
  • UsukiDoll
ok... all we need to is to draw out y=x^2+2x-3 using the same techniques we did and from there we can find the vertex. The vertex of a parabola is the point where the parabola crosses its axis of symmetry.
UsukiDoll
  • UsukiDoll
we can just pick our x to be -1,0, and 1 x y=x^2+2x-3 0 y=0^2+2(0)-3 as before 0^2 means writing 0 twice (two times ) but 0 x 0 = 0 similarly 2(0) means 2 x 0 which is again 0 so we have 0+0-3 which is -3 our first point is (0,-3)
UsukiDoll
  • UsukiDoll
|dw:1434429432668:dw|
UsukiDoll
  • UsukiDoll
it's not drawn to scale... so the graph is a bit ugly...it's just a rough sketch
anonymous
  • anonymous
so the answer is B!!!!!!!!
UsukiDoll
  • UsukiDoll
umm not yet.. for many reasons 1. we don't have enough information 2. we aren't sure if the graph is really above or below the x-axis
UsukiDoll
  • UsukiDoll
we are also not sure if it's on the x-axis or y-axis.
anonymous
  • anonymous
my teacher said to look at this
UsukiDoll
  • UsukiDoll
x y=x^2+2x-3 so if x = 1 y = (1)^2+2(1)-3 = 1+2-3=3-3=0 so the next point is (1,0) if x = -1 y=(-1)^2+2(-1)-3 = 1-2-3 =-1-3=-4 (-1,-6)|dw:1434429719251:dw|
anonymous
  • anonymous
since the example she gave me is +3 and mine is -3 im just gonna go with the opposite of the answer
anonymous
  • anonymous
so im gonna go with b
UsukiDoll
  • UsukiDoll
I don't think that's a great idea. . . in fact.... that's not a good strategy
UsukiDoll
  • UsukiDoll
there is more than one method to solve these problems as long as it doesn't break any math rules... we are using the graphing method to find the vertex which is the lowest or highest point of the graph. In this case, it's the lowest point of the graph which is caused by (-1,-4)
UsukiDoll
  • UsukiDoll
|dw:1434430022195:dw|
anonymous
  • anonymous
so it is b
UsukiDoll
  • UsukiDoll
I need another point to complete this parabola... if I have x = -3 I should have y = 0 I have y = 0 in the right, now I need y =0 in the left y=x^2+2x-3 = (-3)^2+2(-3)-3 = 9-6-3 =3-3=0 |dw:1434430081244:dw|
anonymous
  • anonymous
the answer is b
UsukiDoll
  • UsukiDoll
|dw:1434430156778:dw| yes ... do you know why?
anonymous
  • anonymous
yep
UsukiDoll
  • UsukiDoll
ok, can you explain to me why the answer is below the x -axis ?
anonymous
  • anonymous
ok this next one i already answered i just need u to say if im right or not, k?
UsukiDoll
  • UsukiDoll
alright but can you please answer my question first? I want to see if you understand why it's below the x-axis
anonymous
  • anonymous
because the vertex of the parabola is the point where the parabola crossed its axis of symmetry and that happened under the x-axis. BOOM!!!!!!!!!
UsukiDoll
  • UsukiDoll
yes and also (-1,-4) is the lowest point in the graph... and since the vertex is (-1,-4) it is indeed under the x-axis.. ok we can move on to the next question .
anonymous
  • anonymous
i think the answer is 1
UsukiDoll
  • UsukiDoll
ok this time we have to use the discriminant which is b^2-4ac our equation y=-4x^2+3x+2 is in the form of y=ax^2+bx+c. So we let a = -4, b =3, and c = 2 plugging these values into the discriminant gives us (3)^2-4(-4)(2) 9-4(-8) = 9+32=41 I got a massive number.. I'm reading that if the discriminant is 0, there is 1 solution discriminant is negative -> no solution discriminant is positive -> 2 solutions
anonymous
  • anonymous
so the answer is 2?
UsukiDoll
  • UsukiDoll
I need the quadratic equation to go further.. hold on \[\LARGE \begin{array}{*{20}c} {x = \frac{{ - 3 \pm \sqrt {41} }}{{-8}}} \\ \end{array}\]
anonymous
  • anonymous
kk
UsukiDoll
  • UsukiDoll
\[\LARGE \begin{array}{*{20}c} {x = \frac{{ - 3 + \sqrt {41} }}{{-8}}} \\ \end{array}\] \[\LARGE \begin{array}{*{20}c} {x = \frac{{ - 3 - \sqrt {41} }}{{-8}}} \\ \end{array}\]
UsukiDoll
  • UsukiDoll
the negatives cancel out for the second one \[\LARGE \begin{array}{*{20}c} {x = \frac{{ 3 + \sqrt {41} }}{{8}}} \\ \end{array}\]
UsukiDoll
  • UsukiDoll
\[\LARGE \begin{array}{*{20}c} {x = \frac{{ - 3 + \sqrt {41} }}{{-8}}} \\ \end{array}\] \[\LARGE x = \frac{3}{8}-\frac{\sqrt{41}}{8}\]
UsukiDoll
  • UsukiDoll
the discriminant was positive to begin with... and I don't see any negative signs inside the radical, so we have 2 solutions
anonymous
  • anonymous
thx
anonymous
  • anonymous
can u check this one too?
anonymous
  • anonymous

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