Graph the given quadratic function. Identify the vertex, axis of symmetry, intercepts, domain, range, and intervals for which the function is increasing or decreasing. f ( x ) = 3 x 2 + 12 x f\left(x\right)=3x^2+12x f ( x ) = 3 x 2 + 12 x

Hey everyone, in this problem we want to graph the given quadratic function and then identify all of the elements of our parabola. For the function f of x is equal to 3x squared plus 12 x. Now this is in standard form, so we wanna go ahead and convert this to vertex form by completing the square. So let's go ahead and get started with completing the square. I first wanna identify a, b, and c in my function. So a here is 3, b is 12, and then c is just 0. I don't have one here. So we can go ahead and get started with step number 1, which is to factor a out of my first two terms. So here a is 3. I'm gonna pull an a out of or pull pull a 3 out of both of these terms. So I have x squared and then a plus factoring a 3 out of that 12 x leaves me with 4 x. Now step 1 is done. Moving on to step 2, which is to add and subtract b over 2 a squared inside of our parentheses. So calculating b over 2a first, b is 12. 2 times a a is 3. This leaves me with 12 divided by 6 which is just 2. Now if I take that 2 and I square it, that leaves me with 4. So I'm going to add and subtract 4 inside of my parentheses. So 3 times x squared plus 4x plus 4 minus 4. Now the first part of step 2 is done. Moving on to the second part of step 2 is to move that subtraction outside. So I wanna move this negative 4 and, of course, times my a outside of here. So this is minus 4 times 3. Now that cancels that negative 4 on the inside. So I can go ahead and rewrite this a little bit more compactly. 3 times x squared plus 4x plus 4. And then on the outside, I have this minus 4 times 3, which I can go ahead and simplify to minus 12. Okay. So now that we have step 2 done, moving on to step 3, which is when we need need to factor to x plus b over 2 a squared, and then simplify. Now we already computed b over 2 a, and we got 2, so we're going factor 2, 3 times x plus 2 squared, and then minus 12. So we're done. We don't need to simplify anything else here. We already have our vertex form, so step 3 is done. Moving on to step 4, which is to go ahead and graph from a vertex form. We have our vertex form equation. F of x is equal to 3 times x plus 2 squared minus 12, and I'm gonna copy that right down here so that we can get to graphing. 3 times x plus 2 squared minus 12. Okay. So let's go ahead and move down here and start from step 1, which is to find our vertex. Now we know that in vertex form, this is x minus h and then plus k. So since here I have a positive 2, that tells me that since x minus h, this is really x minus negative 2, I know that h is negative 2. And then k is this negative 12. Make sure you're always paying attention to the sign of both of these values. Now is this a minimum or a maximum point? Well, since this is a positive 3, I know that my parabola is going to open upwards. So this is a minimum point. My vertex is at the very bottom. Now moving on to our axis of symmetry, which is x is equal to h. H here is negative 2, so this is the line x is equal to negative 2. Now moving on to step 34, finding all of our intercepts, we know that we need to do a little bit more work right down here. So let's go ahead and start by computing our x intercepts and setting our function equal to 0. So I have my function here, 3 times x plus 2 squared minus 12 is equal to 0. Now we can go ahead and solve this. 1st, we want to move this 12 over to the other side by adding. Canceling it on this side leaving me with 3 times x plus 2 squared is equal to 12. Of course dividing that 3 over, canceling it out, and leaving me with x plus 2 squared is equal to 12 divided by 3, which gives me a positive 4. Now we can go ahead and apply our square root property and square root both sides here, leaving me with x plus 2 is equal to plus or minus the square root of 4, which is 2. Now I can finally isolate my x and move my 2 over to the other side by subtracting. And then I cancel it out, leaving me with x is equal to negative 2 plus or minus 2. Now we can go ahead and split this into my 2 possible answers here. So this is really negative 2, plus 2 and negative two minus 2 for my two answers. Now negative 2 plus 2 will just give me 0 and then negative 2 minus 2 is going to give me negative 4. So these are my 2 x intercepts, 0 and negative 4, which I can go ahead and fill in on my table up here, 0 and negative 4. Now for our y intercept, we wanna go ahead and plug 0 into our function, leaving us with 3 times 0 plus 2 squared minus 12. And simplifying that, 3 times 2 squared minus 12. 32 squared is going to give me 4, so this is 3 times 4 minus 12. 3 times 4 is going to give me 12, so this is just 12 minus 12 which is just 0. So our y intercept here is 0, which I can fill in on my table. And now I have all of the information I need to go ahead and plot and then connect everything with a smooth curve giving me my parabola. So coming back up to my graph here and starting with my vertex point, we can go ahead and graph that at negative 2, negative 12 right down here. And then x equals negative 2, my axis of symmetry, of course, going right through that vertex point. And then for my x intercepts, 0 and negative 4, I can go ahead and plot those, 0 and negative 4. Now my y intercept is also at 0, and I've already plotted that because it's simply at the origin point. So I don't need to worry about that. But I do have enough points to go ahead and connect with a parabola with a smooth curve. And here is my parabola. Now I wanna put those arrows on the end because it goes on forever. And you notice that this parabola is a little bit skinnier because of that factor of 3 at the front we vertically stretched it. Okay. Now that I have my parabola I can go ahead and identify all of this extra information here. Our domain of course has not changed. It is still negative infinity to infinity, all real numbers. And then with our range, since our parabola is opening upward, I know that it's going to go from my minimum point, which is negative 12, all the way to infinity. That's my range. Now for increasing and decreasing intervals, we wanna look where is our graph going up, where is it going down. So looking at my graph here, from in from negative infinity until it reaches my vertex point at x equals a negative 2, it is going down, so it is decreasing. So that first half of my graph from negative infinity to that vertex point at negative 2, it is decreasing. Then on the other half, of course, split by that vertex, we are going up so our graph is increasing here. So this is from negative 2 all the way to infinity. Now we are finally done with this problem, and you are becoming a pro at graphing parabolas. I'll see you in the next video.

Graph the given quadratic function. Identify the vertex, axis of ...

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Graph the given quadratic function. Identify the vertex, axis of symmetry, intercepts, domain, range, and intervals for which the function is increasing or decreasing. $f\left(x\right)=3x^2+12x$