Hey, everyone. Let's work through this example together. So here we want to graph the given quadratic function identifying all of the possible information about it. Now here we have the function \( f(x) = -\frac{1}{2}(x + 1)^2 + 2 \). So let's get right into graphing. Now looking at the very first thing I want to do, I want to identify my vertex, which when written in vertex form is \( h, k \). Now just a reminder of vertex form, remember it's \( a(x - h)^2 + k \). So identifying our vertex, we need to identify both \( h \) and \( k \). Now here in my function, this is \( x+1 \), and I know in vertex form it's \( x-h \). So whenever we have a plus in that, we want to make sure and be careful and identify this as \( x - (-1) \) so that we know our \( h \) is not positive one it is negative one here. So here my \( x \) value, my vertex is negative 1, and then \( k \) we have this positive 2. So the vertex point is \( (-1, 2) \). Now, is this a minimum or a maximum point? Well, looking back at my function, it has this negative at the front, which tells me that my parabola is going to be opening downward, telling me that I have a maximum point as my vertex. So my vertex is \( (-1, 2) \) and it is at a maximum. Now let's move on to step number 2 and identify our axis of symmetry, which is simply \( x = h \), which we know that \( h \) is negative one. So this is simply the line \( x = -1 \). Now moving on to step 3, finding our \( x \)-intercepts, steps 3 and 4 are going to be a little bit more involved because we need to calculate some stuff, so I'm going to be doing my work for these 2 right at the bottom here. Let's go ahead and set up our equation for step 3 and set up \( f(x) = 0 \). So my function here, \( -\frac{1}{2}(x + 1)^2 + 2 = 0 \). So looking down here at my function, since I have something squared and a constant, I know that I'm going to go ahead and use the square root property. Whenever we have our function in vertex form, we can always solve using the square root property. So I'm going to go ahead and move this 2 over to the other side. So subtracting 2 from both sides leaves me with \( -\frac{1}{2}(x + 1)^2 = -2 \). Now from here, I want to go ahead and cancel out this \( -\frac{1}{2} \), which I can do by simply multiplying both sides by \( -2 \), canceling that \( -\frac{1}{2} \) out and leaving me on this side with \( (x + 1)^2 = 4 \). Now from here, I can go ahead and apply the square root property and simply take the square root of both sides, leaving me with \( x + 1 = \pm 2 \). Now from here, I want to go ahead and move that positive one over to the other side by simply subtracting it, so \( -1 \) on both sides, and I'm left with \( x = -1 \pm 2 \), which I know I can split into my 2 possible answers. \( -1 + 2 \) and \( -1 - 2 \), splitting into my two answers. Now \( -1 + 2 \) gives me a positive one, and then \( -1 - 2 \) is going to give me a negative 3. So here I have my 2 zeros or my 2 \( x \)-intercepts, one and negative 3. So filling that in on my table up here, I have 1 and negative 3 as my \( x \)-intercepts, and we can go ahead and move on to step number 4 and find our \( y \)-intercept by calculating \( f(0) \), plugging 0 into my function for \( x \). So my function here, \( f(0) = -\frac{1}{2}(0+1)^2 + 2 \). Now I can simply compute this by simplifying \( -\frac{1}{2}(1)^2 + 2 \). Simplifying this further, \( -\frac{1}{2} + 2 \). Now I have a fraction and a whole number. We can go ahead and change this whole number into a fraction if that helps you out here. So this 2 is really just \( \frac{4}{2} \). So this is \( -\frac{1}{2} + \frac{4}{2} \), which I found by just multiplying 2 times \( \frac{2}{2} \). Now, since these have a common denominator, I can easily add them together or subtract them since I have a negative here. So \( -\frac{1}{2} + \frac{4}{2} \) gives me \( \frac{3}{2} \), and that is my \( y \)-intercept, \( \frac{3}{2} \). I can go ahead and fill that in on my table as well. \( \frac{3}{2} \). Now it's totally okay to get a fraction for anything, we're still going to be able to plot that. So now I have all of my information that I need in order to graph, and and I can go ahead and move on to step 5 and simply plot and then connect with a smooth curve. So let's go back up to our graph here. Now looking at all of the information I have in my table, I'm gonna first plot my vertex at \( (-1, 2) \), and then my axis of symmetry at \( x = -1 \). I'm going to draw my dotted line right through that vertex point. Now for step 3, I can go ahead and plot these \( x \)-intercepts at 1 and negative 3, positive 1, negative 3, and then finally my \( y \)-intercept at \( \frac{3}{2} \), which is just 1.5, so it's in between 1 and 2. Right here. Okay. So I've plotted everything that I calculated. Now I can go ahead and connect all of this with a smooth curve. Now it's okay if your curve isn't perfect. We know that this is a parabola facing downwards. Your curve might not always look perfect and that's totally okay. We're drawing this by hand. So that's all of our graph. Now we have that, and we can go ahead and find all of this remaining information. So first looking at our domain, we know that the domain of every single quadratic function is always going to be negative infinity to infinity, or simply all real numbers. However you want to write that or however your professor wants you to write that is totally okay here. Both are correct. Now looking to our range, our range here, since we have a maximum point, is going to go from negative infinity until we reach our maximum point, which here is \( y = 2 \) at that vertex. Remember that 2 is included in your range because I only have a point plotted there, so we wanna make sure we use a square bracket on that range. Now we look for our increasing and decreasing intervals. Remember that these are just for what \( x \) values is your graph going up, for what \( x \) values is your graph going down. So looking at our graph here, from negative infinity until I reach that vertex point at \( x = -1 \), my graph is going up, so it is increasing. So from negative infinity to \( -1 \), it is increasing. And then from \( -1 \) to positive infinity, my graph is going down or decreasing. So \( -1 \) to infinity. Now that \( -1 \) is never going to be included in these intervals because it is the point that it switches from increasing to decreasing, but it will always be your vertex point splitting that for our parabolas here. So that's absolutely everything that we need to do for this graph and we found everything we need to. Thanks for watching and let's get some practice.
Table of contents
- 0. Fundamental Concepts of Algebra3h 29m
- 1. Equations and Inequalities3h 27m
- 2. Graphs1h 43m
- 3. Functions & Graphs2h 17m
- 4. Polynomial Functions1h 54m
- 5. Rational Functions1h 23m
- 6. Exponential and Logarithmic Functions2h 28m
- 7. Measuring Angles39m
- 8. Trigonometric Functions on Right Triangles2h 5m
- 9. Unit Circle1h 19m
- 10. Graphing Trigonometric Functions1h 19m
- 11. Inverse Trigonometric Functions and Basic Trig Equations1h 41m
- 12. Trigonometric Identities 2h 34m
- 13. Non-Right Triangles1h 38m
- 14. Vectors2h 25m
- 15. Polar Equations2h 5m
- 16. Parametric Equations1h 6m
- 17. Graphing Complex Numbers1h 7m
- 18. Systems of Equations and Matrices1h 6m
- 19. Conic Sections2h 36m
- 20. Sequences, Series & Induction1h 15m
- 21. Combinatorics and Probability1h 45m
- 22. Limits & Continuity1h 49m
- 23. Intro to Derivatives & Area Under the Curve2h 9m
4. Polynomial Functions
Quadratic Functions
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