Theorem – The lengths of tangents drawn from an external point to a circle are equal – Circles | Class 10 Maths.General and Middle Terms – Binomial Theorem – Class 11 Maths.Arithmetic Progression – Sum of First n Terms | Class 10 Maths.Arithmetic Progression – Common difference and Nth term | Class 10 Maths.Definite Integrals of Piecewise Functions.ISRO CS Syllabus for Scientist/Engineer Exam.ISRO CS Original Papers and Official Keys.GATE CS Original Papers and Official Keys.And so this would be the graph of the line y equals negative one half plus x+5 for x less than 8. It's less than 8 not less than or equal to, I want a little hole here, a little circle showing that I'm not including 8. The domain restriction says graph this line only for x less than 8, so I don't want to go past 8. Well that's the line y equals negative a half x plus 5 but it, it's not correct for the domain restriction. And if I connect those points, I'll get a nice line. You kind of tell by looking at what you would need x to be, right? You'd need this quantity to be -5 so you need x to be 10. In this case 0 5 is the y intercept so that's 3, 0 5 would be about here, and the x intercept I'll get by solving the equation 0 equals negative one half x plus 5. To get a good graph of a line, all you really need to do is find the two intercepts. Now this is just a line so it's pretty easy to graph. I have y equals negative one half x plus 5 restricted to the domain x is less than 8. And that's the entire graph of my quadratic restricted to the domain x between 0 and 5 right? So, I just graph here and in fact I should even erase that point. So I'm just going to graph back to this point. It'll look something like this and notice, I stop here at x=5 because the domain restriction tells me to stop there and then when I graph to the left I don't want to go past this point because this is where x=0. Alright let me put that point here.Īlright, and so now I'm ready to draw my graph. That's about that's about a unit so that's 9. I get 9, so 2 9 is going to be my vertex. I just need to plug 2 into my function, and I get 1+2 or 3, 5-2 times 3. So the vertex is somewhere on this line x=2. So that's going to be 4 over 2 which is 2. So I can just average the two x intercept values -1 and 5 and I'll get the x value of the vertex. Remember if you have the x intercepts, the vertex is going to be exactly halfway between them. And also it's pretty easy to find the vertex. We get that by plugging x=0 in so we get one times 5, 5. And since it's easy enough to find, let's find the y intercept. The x intercept is going to be -1 0 and 5 0, so here's -1 0, here's 5 0. First thing I like to do is to plot the intercepts. Now first of all, quadratics are really easy to graph. Well anyway let's graph this quadratic function and see what the restricted domain does to the graph. You want to make sure that that always ends up positive or that the input is always positive something like that. We would make it a restriction like this if we just wanted to keep it to a certain interval of numbers like between 0 and 5 and this is often done when you're doing mathematical modelling, you're trying to make the function for example model revenue. Now normally a quadratic function is defined for all real numbers. I have a function, a Quadratic function the quantity 1+x times the quantity 5-x, restricted to the domain x is between 0 and 5. I want to talk about domain restrictions.
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