 # 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.

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Thanks for watching and I hope you’ve learned something! 