# Give The Maximum Or Minimum Value The Intercepts And State The Range. Sketch Each Function. A. F

How do I find the range of the function [math]f\left (x\right) =\left (1+x\right) ^2[/math] when [math]-2\le x\le 2[/math]?

f'(x)= 2(x+1)let f'(x)= 0we get x= -1at x = -1 → abs min ,equals 0at x= 2 and -2 → abs max , equals 9so range of f(x) when -2≤x≤2 is [0,9]Orwe can just sketch the graph on [-2,2] and get range of f

How do I find a quadratic equation given 2 points and no vertex?

While it is true that three equations are needed to find the three coefficients, some conditions might help develop a specific equation. However, you are asking “a” quadratic equation.Assuming a vertical axis of symmetry, the equation would be of the form y = ax^2 + bx + c which can also be written as y = a(x - h)^2 + k.Case 1: If the two points have the same y value,then the axis of symmetry will pass through the midpoint of the segment between them. For example, if the given points are (1, 7) and (5, 7), then the midpoint is found by averaging the x coordinates. (1 + 5)/2 = 3. The axis of symmetry is x = 3.Two equations can then be written from the given points(1, 7) 7 = a(1 - 3)^2 + k or 7 = 4a + k(5, 7) 7 = a(5 - 3)^2 + k or 7 = 4a + kAs expected, we have only one equation with two unknowns, resulting in no particular solution. However, once again, looking for a solution, choose a value for k, say -5, the equation becomes 7 = 4a - 512 = 4aa = 3The equation becomes y = 3(x - 3)^2 -5which becomes y = 9(x^2 -6x +9) -5and then y = 9x^2 -54x + 76.Case 2: If the two points have the same y value,Example (1, 7) and (3, 31)7 = a(1)^2 + b(1) + c and 31 = a(3)^2 + b(3) +c(1) 7 = a + b + c and (2) 31 = 9a + 3b + cSubtracting equation 1 from equation 224 = 8a + 2bdividing by 212 = 4a + bb = 12 - 4aLet a = 2 or any other chosen valueb = 12 - 4(2)b = 4Substituting into equation 17 = (2) + (4) + cc = -1Answer: An equation going through the two points is y = 2x^2 + 4x - 1

How can I graph an exponential function on excel?

To plot an exponential function, what you can do is type in your function. Let’s say for example your function is [math]y = 5^x[/math]. What you can do is create your range for the x-values. Type in a header for your range (just call it x). Let’s say we will have these values from 0 to 30 (you can give them any minimum or maximum you like).After that, what we can do is create one more header, called 5^x. This will be the range for our y-values. What I like to do is use an Excel table for this kind of thing. The reason why is when you use Excel tables for mathematical formulas, Excel does the heavy-lifting for you and copies the formula all the way down the table.To create the Excel table, highlight the x-values column with its header, and then highlight the next header over along with the empty range underneath it. Press Ctrl+T. You will see a window that looks just like this:Click OK. Now in your 5^x formula, in the first empty under your header, type in the formula like this:=power(5,b3)

The =power() function tells Excel to raise 5 to each successive value in the preceding column (the x-values column). Press Enter. Excel does the rest.Highlight the entire table and then go to INSERT. Click on SCATTERPLOT and then choose the smooth curved line option. It will give you a curve that is for the most part very flat (this is because the nature of the growth of 5^x.If you want its curvature more pronounced, while both ranges are highlighted, you can click on the bottom right hand corner of the selection and drag it up. This will readjust the graph to make its curvature a little more pronounced.For example, in this screenshot, I adjusted the selection range to just account for the first 10 values of the function.One more thing before I end this entry. I suggest that SCATTERPLOT be used for two reasons: (1) if you use the LINE GRAPH instead, you will get this instead:If you look closely, you can see that there is a horizontal line at the bottom of the graph. Technically it is not a horizontal line because the x-values are still increasing, but its hard to demonstrate that at the rate the y-values are growing.(2) We instead have two graphs. The x-value range and y-value range are being treated as two different data sources. Not only this, the graph looks very rigid. SCATTERPLOT treats both ranges as inputs and outputs of a function.Hope this helps. :)

A function with domain all real numbers satisfies all of the following conditions:?

1. This one is simple. You know this much:

f(-inf) = -inf

f(0) = 2

f(3) = 0

f(5) = 2

f(inf) = -inf

There are three zeroes. It crosses 0 on it's way to f(0). It touches 0 at f(3). And it crosses 0 at some point after f(5).

You know one is at x=3. All you can say about the other two is that one is before x=0, and one is after x=5; that's as accurate as you can get.

2. Because it is continuous from -inf to inf, there are no vertical asymptotes. And because it has the points (-inf,-inf) and (inf,inf), there are no horizontal asymptotes.

In other words, NO asymptotes are in this function.

3. There are two local maxima, at the tops of the hills (which are at the points (0,2) and (5,2)). There is a local minimum at the bottom of the ditch at the point (3,0).

4. Use the above answers to draw it. Graph those points you know, remember where it increases and decreases, and smooth it out. There should be a hill that tops out at (0,2), and a ditch that bottoms at (3,0). The graph should look like an M.

Edit: Oops. I made one reading mistake, and it threw off half of my answers. Sorry!

The user below me is correct. The problem does state that the graph decreases in the range [5, infinity). This changes A LOT. It creates a local maximum at (5,2), it changes the shape from an N to an M, it adds a root, and it changes one of those f() values I listed in #1.

Man, all from one reading mistake.

I corrected those things in my post. I also realized that you cannot use the Solver on those unknown roots, because you need a formula for that. I fixed that too.

Thanks for pointing that out.

I see mistakes happen often on this site. My advice? Don't always take people's word on their answers, especially this early in the school year, when most of us haven't transitioned back in to school mode yet. As I'm going around answering, I see that almost 75% of the answers have mistakes in simple arithmetic that throw off everything. I even saw somebody say 9x8=18.

How I find the turning point of a quadratic equation?

The turning point is called the vertex. There are a few different ways to find it. Fortunately they all give the same answer.You’re asking about quadratic functions, whose standard form is [math]f(x)=ax^2+bx+c[/math]. When [math]a \ne 0[/math], these are parabolas. We know [math]f(x)[/math] has zeros at[math]x = \dfrac{-b \pm \sqrt{b^2–4ac}}{2a}[/math]We also know the vertex is right in the middle between the two zeroes. If we add up the two solutions to find the average, the [math]\pm[/math] part goes away and we’re left with:[math]x = -\dfrac{b}{2a}[/math][math]y = f(-\frac{b}{2a} ) [/math]Another way to see this is the vertex is the point where the function is flat, i.e. where its slope or derivative is zero. The derivative [math]f’(x)=2ax+b.[/math] So [math]2ax+b = 0[/math], or [math]x=-\frac{b}{2a}.[/math]The last way is by completing the square:[math]ax^2+bx+c = a(x^2 + \frac b a x + \frac c a)=a( (x+\frac{b}{2a})^2 + \frac{c}{a} - \frac{b^2}{4a^2}) = a(x+\frac{b}{2a})^2 + c - \frac{b^2}{4a}[/math]When the expression in parentheses is zero, the square will have its smallest value, so this will be the minimum or maximum (depending on whether [math]a[/math] is positive or negative). Either way it’s the vertex. So[math]x+\dfrac{b}{2a} = 0[/math][math]x = -\dfrac{b}{2a}[/math]Three different ways, same answer.

Find the range of f(x) =2-3x, x belongs to real numbers, x>0?

To find the range of the function f(x)= 2 - 3x, let's take f(x) = yi.e. f(x) = 2 - 3x = yi.e. 2 - y = 3xor x = 2 - y 3As x > 0, so 2 - y > 0 3 i.e. 2 - y > 0 or y < 2Hence range of this function is ( - infinity, 2)

What are the increasing and decreasing intervals in a parabola?

I assume you're asking about a quadratic function [math]f(x)=ax^2+bx+c[/math] since the graphs of quadratic functions are parabolas. There are other parabolas whose axes of symmetry aren't vertical but they are not the graphs of functions.The vertex of such a parabola occurs when [math]x=-\dfrac b{2a}[/math]. On one side of that value, the function will be increasing and on the other it will be decreasing. If [math]a[/math] is positive, then the parabola opens upward, so the function decreases on [math]\left(-\infty,\frac{-b}{2a}\right][/math] and increases on [math]\left[\frac{-b}{2a},\infty\right)[/math]. But if [math]a[/math] is negative, then just the reverse. Image source: new.edu College Algebra

What is stopping potential in photoelectric effect?

DefinitionThe stopping potential is defined as the potential necessary to stop any electron (or, in other words, to stop even the electron with the most kinetic energy) from reaching the other side.What happens in photoelectric effect is that, an incident radiation strikes the surface of the metal. Now, as the incident photons with a certain energy, strike the surface of the metal, they transfer their energy to the electrons in the outer shell of the metal atom so that they ionize and come out of the metal atom. This is called emission and this happens only when the incident photon strikes with sufficient energy. The energy should be sufficient for the electron to be able to come out of the metal atom. However, the energy that this electron receives may either be exactly the same amount as required by it or it may be of some amount which is slightly greater than what is required. In the latter case (lets say case 2), the electron uses the extra amount of energy as its kinetic energy and thus gains some speed. However in the former case (lets say case 1), the electrons are just emitted but they gain no speed. The entire amount of energy is used in the emission. So, in the tube, if the photocurrent is to be detected, the requirement is that some current flows through the circuit. For this, electrons must flow. And for that, electrons should gain some speed.Explaining stopping potentialWhen we apply a positive potential on the plate placed directly opposite to this metal plate, the plate attracts the negative electrons. A force acts on these electrons and thus they get accelerated towards this plate. So, the more positive potential we apply on this (collector) plate, more rapidly will the electrons flow through the circuit. But if we apply a negative potential on this (collector) plate, the electrons will not approach so easily. Only the electrons with sufficient energy and hence, velocity, will be able to surpass this opposing potential called retarding potential of the (collector) plate. If we keep on increasing this negative (retarding) potential, there will be one point when no more electrons reach this plate and thus, no electron flows through the circuit. Thus, current flow through the circuit stops. This potential applied on the (collector) plate is called as the stopping potential.

How does one find the intersect point between 2 trendlines in Excel?

You can use solver to find such points, but firstly you need to setup a sheet for that and calculate coordinates of the two equation within excel cell. For that you will need the two formulas: INTERCEPT() and SLOPE(), The first one will give you the constant of equation while the slope gives you the coefficient of x in the equation.You can find the following formulas in the attached sheet in the cells B17, B18 and C17 and C18.After that you need to have process of trial and error where you need to find a point where both equations have same value of x. For that you need to setup a Solver Problem. What i have done is to that for an intersecting point, the value of x should be same for both of them!See the below setup of Solver, you can run for the values in cell D20 to get the desired result.Once you are done with the x coordinate you can find y-coordinate by substituting it in either of the equations. See this worksheet for your convinience:Trendline.xlsx