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 ) = − ( x − 5 ) 2 + 1 f\left(x\right)=-\left(x-5\right)^2+1 f ( x ) = − ( x − 5 ) 2 + 1

In this problem, I want to graph the quadratic function and then identify all possible elements of my parabola. So the function I have here is f of x is equal to negative x minus 5 squared plus 1. So let's go ahead and jump right into graphing and identify our vertex h comma k. Now because we know that this is x minus h, our h value is 5, and then our value for k is 1, so this vertex is 5 comma 1. Now, is this vertex a minimum or a maximum point? Well, since I have a negative in front of my function, that tells me my parabola is opening downward, which means my vertex is at a maximum at the very top of that parabola. So 5 comma 1 and a maximum. Now to identify our axis of symmetry at x is equal to h, I know that h is 5, so this is simply the line x is equal to 5. Now for steps 34, we know that we need to do some calculations, which I'll be doing right down here. And first, for step 3, we want to solve f of x is equal to 0 to get our x intercepts. So taking my function negativex minus 5 squared plus 1 and setting that equal to 0, I can go ahead and solve this equation. I'm gonna use the square root property here as I do for all of my quadratic functions in vertex form. So let's first move this one over to the other side by subtracting. Leaves me with negative x minus 5 squared is equal to negative one. Now I want to get rid of this negative on the outside here, which I can do by simply dividing both sides by negative one. Canceling that out and leaving me with x minus 5 squared is equal to negative one divided by negative one, which is a positive one. So now I can go ahead and apply my square root property by taking the square root of both sides, leaving me with x minus 5 is equal to plus or minus the square root of 1, which is just 1. Now I can go ahead and isolate x by adding 5 to both sides, leaving me with x is equal to 5 plus or minus 1. Now here, we wanna go ahead and split this into our 2 possible answers, 5 +1 and 5 minus 1. So I have 5+1 and then 5 minus 1. Now computing those to get our final answers, 5+1 is 6 and 5 minus 1 is 4 and these are my x intercepts, 6 and 4, which I can go ahead and fill in on my table up here, 6 and 4. Now to compute our our y intercept by simply plugging 0 into our function, we get negative 0 minus 5 squared plus 1, and that is equal to negative. 0 minus 5 is still negative 5 squared plus 1. Don't drop this negative. It's still there. Negative 5 squared, though, does give me a positive 25 plus 1. So I'm left with negative 25 plus 1, which is going to give me negative 24, and that is my y intercept, a little bit of a larger y intercept here, negative twenty 4. Okay. Now I have all of this information. I can go ahead and plot all of this on my graph and then connect with a smooth curve. So let's go back up to our graph here and plot everything that we have starting with our vertex at 5 comma 1. So putting that right here, and then plotting my axis of symmetry right through that vertex point at x equals 5. Then my x intercepts at 46, which are right here. And then my y intercept is actually a negative 24. And my graph is not scaled in a way that I can go ahead and plot this because it would look a little bit crazy. So it's totally okay if you can't always plot your y intercept. Sometimes it's gonna be a little too far on your graph in order to plot it but we still have a good idea of what the graph of our function looks like so that we can go ahead and connect it with a curve. So we know that it is a downward facing parabola with these three points. So I'm going to go ahead and connect these, of course, putting arrows on either end because it keeps going forever. Now that I have my graph, I can go ahead and identify all of this remaining information starting with our domain which for every quadratic equation is going to be the same and it is a negative infinity to infinity. Now for our range, since we have a maximum point at the very top of our parabola, our range is going to go from negative infinity until that maximum point, which here is at y equals 1, and of course including 1 in a square bracket. Now finally, for our increasing and decreasing intervals, remember increasing is where our graph is going up, decreasing is where it's going down, and we're looking at x values here. So for our increasing, looking at this side, from negative infinity until I reach that vertex point at x equals 5. We are increasing because our graph is going up, so negative infinity to 5, and then decreasing on the other side of my parabola as it falls right back down. So from 5 to infinity. And that's it. I know that these are long problems, but you're doing a great job. Let's get some more practice.

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. $f\left(x\right)=-\left(x-5\right)^2+1$