Guide :

# write the equation for the function having a graph that meets all the conditions

has the same shape as graph g(x)=-3x^2 and has a maximum value at (-2,4) a f(x) = 1/3(x-2)^2+4 b (fx) = -3(x-2)^2+4 c f(x) = 3(x+2)^2-4 d f(x) = -3 (x+2)^2+4

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## Suggested Questions And Answer :

### write the equation for the function having a graph that meets all the conditions

What does g(x) look like as a graph? It's a good idea to draw a picture to help you visualise the answer to this question. When we show g(x) as a graph it looks like an inverted U with it's maximum point at (0,0). The graph never moves any further into the positive region of the vertical axis, that is, g(x), but it spreads itself symmetrically because the square of a negative number is the same as the square of the equivalent positive number (for example, -1 and +1 have the same square =1). This makes the vertical axis look like a mirror. Now to the question. The spread of the arms of the U is governed by the constant -3. The answer to the question must preserve the spread so we can reject option a, because that would be a narrower shape than g(x). Next we have to see which option puts the maximum point at f(x)=4. The maximum for g(x) is at (0,0) the "top" of the upturned U. We need to slide the U upwards, more positive, so we can reject option c, because it would force the curve downwards, more negative. We're left with options b and d. When f(x)=4 we know we must have x=-2 to satisfy the requirements, so we substitute this value into options b and d and for b we get a number which is far too negative, while option d gives us 4, so d is the right answer!

Let's illustrate what function notation is by starting with an equation y=3x+2. We know this equation can be represented by a straight line graph. If I say what is y when x=1? You would work out that y=5. We can say that y is a function of x, that is, multiply x by 3 then add on 2. The equation is a short way of expressing this. In the same way we can use f(x), that is, a function of x, or f of x, where f(x)=3x+2. Instead of asking what is y when x=1, we can write f(1)=? It's shorthand. But it's more than that. Suppose we write z=2y-1 and y=3x+2 and ask the question what is z when x=1? First you work out y, which we know is 5, then we would put y=5 into z=2y-1 and z would be 9. Using function notation and writing g(y)=2y-1 is the same as z=2y-1. But here's the clever bit: we can also write g(f(x)) or gof(x) or g of f(x) to mean the same. Shorthand again. gof(1)=9. We would also have written g(x)=2x-1 instead of g(y)=2y-1. This shorthand function notation means we won't run out of letters like x, y, z, because we can simply write our functions in terms of x and leave it at that. And you can write things like f(x)+g(x) or h(x)=(f(x)+g(x))/(f(x)-g(x)). It's a whole new world! You still need to work out the answers with as much effort as before, but it's easier to express the question. And in some cases there are quicker ways to get the answer. I hope this helps.

### Graph Square-Root Functions State the domain and range for each of your equations. y=2√x+1

square root (x)..portant thang is  x>=0 the +1 shift axis up1...parallel tu x-axis it look like parabola with horizontal axis, point tu rite

### let f(x)=x^105+x^53+x^27+x^13+x^3+3x+1.if g(x) is inverse of function f(x),then g'(x) is

let f(x)=x^105+x^53+x^27+x^13+x^3+3x+1.if g(x) is inverse of function f(x),then g'(x) is ?   We have f(x) = y = x^105 + x^53 + x^27 + x^13 + x^3 + 3x + 1   Normally, the inverse function is given by f^(-1)(x) = g(x), where g(x) is gotten by solving the equation y = x^105 + x^53 + x^27 + x^13 + x^3 + 3x + 1 for x, giving x = g(y). We then write y = g(x) and give the inverse function as f^(-1)(x) = g(x). However, we cannot solve y = x^105 + x^53 + x^27 + x^13 + x^3 + 3x + 1, for x. Instead we get the inverse by writing x = y^105+ y^53 + y^27 + y^13 + y^3 + 3y + 1 = h(y). If we plot the function x = h(y), we will be plotting the inverse function f^(-1)(x) = y. If we were to plot both f(x) and h(y) (i.e. f^(-1)(x)), we would see that they were reflections of each other about the line y = x. We would also be able to see what intersection points, if any, there were between the two curves. The question asks us to find g’(x), the slope of the inverse function, which happens to be:  h(y) = x = y^105+ y^53 + y^27 + y^13 + y^3 + 3y + 1 Differentiating, dx/dy = 105y^104+ 53y^52 + 27y^26 + 13y^12 + 3y^2 + 3 Therefore, g’(x) = dy/dx = 1/(dx/dy) = 1/(105y^104+ 53y^52 + 27y^26 + 13y^12 + 3y^2 + 3) The solution options given for the slope are numerical values, therefore we must be getting asked for the slope at a particular point, which is some value for x and some value for y. The options for the slope are 3, 1/3 and -1/3. Looking at the expression for g’(x) = 1/(105y^104+ 53y^52 + 27y^26 + 13y^12 + 3y^2 + 3) The only viable option is to choose y = 0, giving g’(x) = 1/3 Answer: g’(x) = 1/3   Here is an image of f(x) and h(y) plotted on the same graph.     The blue line is: f(x) = y = x^105 + x^53 + x^27 + x^13 + x^3 + 3x + 1 The red line is: h(y) = x = y^105 + y^53 + y^27 + y^13 + y^3 + 3y + 1 The white line is: y = x. You can see that both curves are reflections of each other about the line y = x. They have a point of intersection at approx. (-0.5, -0.5) The intersection of h(y) on the y-axis is at about y = -1/3. The intersection of h(y) on the x-axis is at x = 1. These values give an approximate slope, in that area, of 1/3. This would confirm the value of 1/3 found above for g’(x) at y = 0.

### Linear word problem help.

We can call x the number of candy bars sold. The linear equation for the fund can be represented by y=150+3x dollars (this is the rule). That's the initial amount of \$150 plus \$3 for each candy bar. When y=\$237, the goal, 237=150+3x, so 3x=237-150=87 and x=87/3=29 to meet the goal. A graph of y=150+3x can be drawn where 150 is the y intercept and -50 is the x intercept. By joining these intercepts and continuing to the right you will see a graph showing values of x and corresponding values of y on the graph to the right. The graph can be drawn near the bottom of the page to allow y to extend to about 240 with suitable scaling. The only reason to draw the graph away from the left of the page is to accommodate x=-50, the x intercept. The only meaningful part of the graph is to the right of the y axis and above the x axis, the positive region for both x and y. If you read off the x value for y=237 you should see that it's 29. You can also see what value of y corresponds to x=14. It should be y=192. A comfortable scaling for both axes is tenth of an inch for each unit, so 80 units on the x axis is 8" and 240 units on the y axis is 2 feet (a long sheet of paper!). To shorten the y axis, you can take y from 150 to 240, 90 units, or 9". That means losing 5" on the x axis so that x starts at 0 and goes up to about 30 or 3". You don't need the same scale on each axis, so you could divide the y axis so that 1/10=2 units, making the y axis 12" instead of 24". Play around with the scaling and show your daughter what you are doing. To make a table of x and y values, for example: x   y 0   150 10 180 20 210 30 240 Join these points to get the line and choose the scale that gives you the best readings. I hope this helps you both.

### how to make an equation and draw a graph without a scale given.

yer long list av werds inklude "water freez & water boil" that tell yu teech want skale tu go from freez tu boil 0 tu 100 sentigrade & 32 tu 212 in farenhite