# How to Sketch a Parabola – Example 3 (y = x^2 + 6x + 10)

Let’s do a quick 3rd and final video on how to sketch a parabola or quadratic function. And the function that I have here is: y = x^2 + 6x + 10. I’m going to make things a bit clearer by labelling our coefficients. So I label the x^2-coefficient:

(a);
our x-coefficient: (b);
and our free term: (c).

And I’ll note that the coefficient in front of the x^2 is actually a positive

1. And as we’ve found in past videos, we only need 4 pieces of information to be able to sketch a parabola. Number 1: we need the concavity, which is determined by our (a) coefficient… and we see that it’s a positive number, so that means our parabola is going to “smile”… or if you like, open upwards. The second bit of information we’re looking for is the y-intercept. And that’s just going to be our (c) value, or more formally when x = 0, that’s positive 10… so let’s mark that on our chart… 10. And note that I’m not going to draw this to scale – this is just a sketch.

Number 3, we need is the turning point, or the vertex. And that’s given by the formula: The x-coordinate is: -b/2a And the y-coordinate is: -[b^2-4ac]/4a … Alright, so substituting our numbers in… -6/(2*1) for ‘x’ And negative of 6^2, which is 36… minus 4x1x10, which is 40… divided by 4×1. So the first term is going to evaluate to -3 And for our y-coordinate, 36-40 is negative 4. So that’s minus negative 4 divided by 4 And that’s going to yield us: (-3, 1) So the x-coordinate of our turning point is -3 and the y-coordinate is positive 1.

So our turning point is located here. And the 4th and final part is to find the x-intercepts. Well, I know straight away from looking at this, we’re not actually going to get any x-intercepts at all. First of all, visually: the y-coordinate of our turning point is positive 1, and the y-intercept is at positive 10 and we know that the parabola opens upwards. So if we are going to get any x-intercepts, it looks like we have done something very wrong.

Ok, the other tell-tale sign is the evaluation of (b^2-4ac), and that comes out to be a negative number. And to find the x-intercepts, we can use the quadratic formula. So x = [-b +/- sqrt(b^2 – 4ac)]/2a Well if b^2-4ac is negative, we won’t get any real answers because the square-root of a negative number can’t be calculated. So in this case, we will not have any x-intercepts, which makes it a little bit harder to sketch, but none-the-less, we can still sketch it. We keep using our artistic skills to try and draw a smooth line across the y-axis and try and make our parabola as symmetrical as possible… (well, that’s not very good, but possibly the best that I can do with a mouse on a primitive program) So I hope this quick video tutorial has helped you.

If you are a Math student, please feel free to subscribe for future videos that may help you on assignments or exams. If you have any question that you would like me to do a video on, please feel free to comment on any of the videos that you’ve seen.
Thanks for watching and I hope you’ve learned something!

#### 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.
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