Find domain and range?
find the range of the function -10x^2-4x+18 also, find the domain of this function f(x)=log base 7 (8-2x) ( i know you can use the rule and change it to (log8-2x)/log7 but idk what the domain is )
Algebra, domain and range help?
i have to find the domain and range for 3 different functions; linear parabola abs. value i know that the domains are your x's and your ranges are your y's, but my teacher confused me when she said put your domains and ranges like 4
How to find domain and range of sqrt(4-x^2) ?
how to find domain and range of sqrt(4-x^2) ? I am confused because sometimes something is infinity and stuff... Rather than just getting the answer, I'd like to know all the steps to the answer. Thank you :)
Find the Domain and Range of the function?
17.) f(x) = 3-2x^2 Domain is all real numbers. The graph of this is a parabola opening downward. It's maximum value is 3 (this occurs at x=0). So the range is (-inf, 3). You can view the graph here: http://www.wolframalpha.com/input/?i=+3-... 18.)f(x) = sqrt (2x^2-1) The radicand must be >= 0, so 2x^2-1 >= 0. The graph of 2x^2 - 1 a parabola opening upward. The graph is above the x-axis (meaning 2x^2-1 >= 0) when x is to the left of (-sqrt2)/2 and to the right of (sqrt2)/2. So the domain is (-inf, (-sqrt2)/2) U ((sqrt2)/2, inf). The sqrt of any nonnegative real number is always greater than or equal to 0. The range is [0, inf). You can view the graph here: http://www.wolframalpha.com/input/?i=sqr... **Disregard the red line. 19.) h(x) = sq. root of 36-x^2 The radicand must be >= 0, so 36-x^2 >= 0. The graph of 36-x^2 a parabola opening upward. The graph is above the x-axis (meaning 36-x^2 >= 0) when x is between -6 and 6 inclusive. So the domain is [-6,6] The sqrt of any nonnegative real number is always greater than or equal to 0. The range is [0, inf). You can view the graph here: http://www.wolframalpha.com/input/?i=sqr... **Disregard the red line. 20.) g(x) = l x+5 l The domain is the set of all real numbers. The absolute value of any number is always greater than or equal to 0. So the range is [0,inf). You can view the graph here: http://www.wolframalpha.com/input/?i=abs%28x%2B5%29
Find the domain and range of the function f(x)=12^2-18?
The domain includes any number you can plug in for x. In this case, this includes any real number (-infinity, infinity). Because x is squared, you are graphing a parabola. Any number, positive or negative makes the first portion (12x^2) a higher positive number. Since you can plug any number in, a greater absolute value of x means a greater value for f(x). The greatest absolute values from the domain, gives you infinity for the range. At x=0 (the vertex of the parabola) f(x) is at it's minimum value. f(0)= 12(0)^2-18 =-18. So, the range is [-18, infinity).
What is the graph and what are the domain and range?
You have a point (2.25, 800) and a slope of 40/-0.25, or -160. The slope is a rate of change - if you lost 40 customers per $0.25, then your slope is 40/-0.25. To find the equation of the line, use the point-slope formula: y - y1 = m(x - x1) y - 800 = -160(x - 2.25) y - 800 = -160x + 360 y = -160x + 1160 The graph will be a line starting at (0, 1160) and going down with a slope of -160. The domain will start at 0 - since the x-axis represents money, you have to exclude negative numbers. To find the rightmost value for the domain, find the x-intercept: 0 = -160x + 1160 160x = 1160 x = 7.25 So your domain is [0, 7.25]. The range will be [0, 1160] - 0 for no customers, 1160 from the y-intercept.
Parabolas and Geogebra?
The vertex form of the equation of a horizontal parabola is given by , where (h, k) is the vertex of the parabola and the absolute value of p is the distance from the vertex to the focus, which is also the distance from the vertex to the directrix. You will use GeoGebra to create a horizontal parabola and write the vertex form of its equation. Open GeoGebra, and complete each step below. Mark the focus of the parabola you are going to create at F(-5, 2). Draw a vertical line that is 8 units to the right of the focus. This line will be the directrix of your parabola. What is the equation of the line? Construct the line that is perpendicular to the directrix and passes through the focus. This line will be the axis of symmetry of the parabola. What are the coordinates of the point of intersection, A, of the axis of symmetry and the directrix of the parabola? Explain how you can locate the vertex, V, of the parabola with the given focus and directrix. Write the coordinates of the vertex. Which way will the parabola open? Explain. How can you find the value of p? Is the value of p for your parabola positive or negative? Explain. What is the value of p for your parabola? Based on your responses to parts iii and v above, write the equation of the parabola in vertex form. Show your work. Construct the parabola using the parabola tool in GeoGebra. Take a screenshot of your work, and paste it below. Once you have constructed the parabola, use GeoGebra to display its equation. In the space below, rearrange the equation of the parabola shown in GeoGebra, and check whether it matches the equation in the vertex form that you wrote in part vii. Show your work. Write the equations of these two parabolas in vertex form: • focus at (4, 3), and directrix x = 2 • focus at (2, -1), and directrix x = 8 • focus at (-5, -3), and directrix y = -6 • focus at (10, -4), and directrix y = 6 How do I do this? Can somebody walk me through this?
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
Find the function f(x) whose graph is obtained by shifting the parabola y=x^2 three units to the right?
find the function f(x) whose graph is obtained by shifting the parabola y=x^2 three units to the right and four units down. also define f(x) to be the larger of x and 2-x. What is the domain and range when you sketch it? express f(x) in terms of absolute value funtion