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Graphing Radical Equations using a Table - Problem 1 2,670 views

Teacher/Instructor Alissa Fong
Alissa Fong

MA, Stanford University
Teaching in the San Francisco Bay Area

Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts

You can always make graphs in Math class using a table of values. Sometimes it’s a little slower but it's usually on of the most accurate. So let’s go ahead and look at this problem where we are asked to find the domain of f(x) equals the square root of x plus 4 and sketch your graph using a table of values.

Okay so let’s talk about the domain. You know that the domain has to be values that make your radicand greater than or equal to 0. Radicand is the stuff under the square root. Be really careful it’s only the stuff under the square root, we don’t care about the stuff that’s outside the square root.

X plus 4, my radicand has to be greater than or equal to 0. So my domain is going to be all x values that are greater than or equal to -4, that’s my domain. So what that tells me is that when I’m creating my table, I don’t want to start my table like -7, 6, 5, 4 something like that, my table should start with -4.

Pick a few points then you're going to substitute them into your function one at a time and find your corresponding y values. I already did that ahead of time but I'll show you the first couple. If I put in -4 right there then I’ll have the square root of -4 plus 4 or the square of 0 which is 0. If I put in -3 I’ll have the square root of 1 which is 1. If I put in -2 I’ll have the square root of 2 which is approximately 1.4, if I put in -1 I’ll have the square root of 3 which is 1.7ish, and if I put in 0 I’ll have the square root of 4 which is 2.

Okay you can verify that on a calculator to get these decimal approximations. I’m ready to put these guys on the graph so here we go. I’m going to draw my graph which I think it’s going to look like the shape of a parabola kind of on its side. My first point was (-4,0) then I had (-3,1), (-2,1.4), (-1,1.7) and then (0,2). My parabola shape is on its side it goes like this and it's curvy. Please be careful again don’t stick an the arrow on that side. That’s like a dead end, that’s the last point or the smallest x value I could use in my domain. If I chose x number smaller than -4 I’d have non-real solutions for my output.

So the shape of my graph looks like this what I did was find my domain before I made my table because that told me what x numbers to input to get my corresponding y output values.

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