 # How to Graph a Parabola with no x-intercepts: y = x^2 + 2

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Here we’re asked to graph a quadratic function, the first step is to determine the y intercept the y intercept occurs when x is 0 so when x is 0 here y you can easily see should be 2 the next step is to determine the x-intercepts in this case you find the x-intercepts when y is equal to 0 so 0 equals x squared plus 2 subtracting 2 and solving this quadratic we can extract the root here don’t forget the plus or minus when you do that but notice what happens here I’m left with plus or minus I times the square root of 2 ok when the x-intercepts turn out to be imaginary there’s that’s telling me the math is telling me there’s no X intercepts all parabolas have a vertex so the vertex while the x-value of the vertex can be gotten with the formula negative B over 2a in this particular case notice a is 1 B is 0 and C is 2 so when I plug in 0 negative 0 over 2 times a well 0 divided by anything is just 0 so when x is 0 the corresponding Y value would be 2 0 squared is just 0 so 0 squared plus 2 is 2 so the vertex is a 0 comma 2 notice here the vertex is the same as the y-intercept you’ll see later when we have our transformations here we have a regular a basic graph here parabola shifted up 2 units so we only really have one point you know one point is not enough to graph a parabola so at this point we’re sort of stuck let’s go ahead and make a chart of values and let’s get a couple more here.

So let’s say we have negative one we already know if we plug in 0 we’ll get – and let’s look at what happens when X is 1 so when y is negative 1 substituting that in negative 1 squared + 2 is 3 okay so the score corresponding Y value there would be 3 now let’s do the same thing by plugging in when we plug in 1 there 1 squared is 1 and there I get 3 as well ok so here’s 3 points at least we know the parabola also opens up because the leading coefficient is positive so let’s graph this okay so 0 comma 2 is the y-intercept and also the vertex 1 comma 3 and negative 1 comma 3 so here’s the parabola you can sort of see it here where the line of symmetry is the Y axis let me go ahead and connect the dots well here you go this is the best I can do on the computer here you notice the parabola opening up there’s my y intercept which also happens to be the vertex at 0 comma 2 and no x-intercepts notice here the math told us that right we got here an imaginary x intercept so it’s telling me no x-intercepts in this case so there it is y equals x squared plus 2 . ## About Post Author

#### Professor Cram

Professor Lawrence Cram is a Visiting Fellow at the Australian National University working in the Department of Applied Mathematics. His interests include astronomy, mathematics, engineering, computing, and physics. Due to his extensive expertise, professor Cram has worked as a Professor of Physics at the University of Sydney and as the Deputy Vice-Chancellor at ANU during 2004-2012. In 2013, he retired as a Master, University House and Graduate House. In January 2014, he was appointed as an acting Deputy Vice-Chancellor at Charles Darwin University. Professor Cram is also a Fellow at the Royal Astronomical Society and the Australian Institute of Physics.