2017年10月19日 · The graph is of general shape #uu# as the #x^2# term is positive Given: #y=ax^2+bx+c color(white)("d")->color(white)("d")x^2+x+1#

2014年9月13日 · The graph of f(x)=x^2 is called a "Parabola." It looks like this: One of the ways to graph this is to use plug in a few x-values and get an idea of the shape. Since the x values keep getting squared, there is an exponential increase on either side of the y-axis. You can see this by plugging in a few values: When x=0, f(x)=0 x=1, f(x)=1^2=1 x=2,f(x)=2^2=4 x=3, f(x)=3^2=9 x=4, f(x)=4^2=16 The ...

2018年7月2日 · See answer below Given: f(x) = x^2 + 1 You can use point plotting. Select any x value and then calculate the y value and plot. ul(" "x" "| " "y" ") ""-4" "|" "17 ""-2" "|" "5 " "0" "|" "1 " "2" "|" "5 " "4" "|" "17 or you can realize that x^2 + 1 is a vertical shift up 1 of the parent function f(x) = x^2 or you can just graph the function using a graphing calculator. TI …

2016年3月23日 · Let y = f(x) and solve for x. We find that there is no inverse function unless the domain of f(x) is restricted. Suppose f(x) = x^2-1 To attempt to find an inverse function, let y = f(x) and solve for x in terms of y... y = x^2-1 Add 1 to both sides to get: y + 1 = x^2 Transpose and take square root of both sides, allowing for either sign: x …

Multiply both sides of this equation by -1 to get: x^2 = 1 Take the square root of both sides of this equation to get: x = +/- 1 Those should be the x-axis crossing points. You could also have factored the equation of -x^2 + 1 = 0 to get: (-x+1) * (x+1) = 0 When either of these factors = 0, the equation is grue, so you set each of the factors ...

2018年4月25日 · f^(-1)(x)=+-sqrt(x+1) Let y=f(x). Then, y=x^2-1 To determine the inverse function, switch the places of x and y, and subsequently solve for y: x=y^2-1 y^2=x+1 y=+-sqrt(x+1) Since this is the inverse, y=f^(-1)(x): f^(-1)(x)=+-sqrt(x+1)

2018年5月15日 · f(g(x))=1/x^2+1 f(x)=x^2+1 g(x)=1/x In order to find f(g(x)), we must substitute g(x) in for x for the equation f(x). (1/x)^2+1 Now, all we need to do is simplify to find the final answer. 1/x^2+1

Add and subtract that number, that is add + 9 - 9 after the 6x: Factor the first three terms: If you followed the 3 steps above both factors are the same and it can be written as the square of a binomial: Distribute to remove the outer (large) parentheses, leaving the square of the binomial intact: Combine the remaining terms as -10 That's in ...

2018年3月4日 · (f@g@h)(x) = 4x^2-8x+5 Given: { (f(x) = x^2+1), (g(x) = 2x), (h(x) = x-1) :} One way of thinking about these function compositions is to go back and forth between the symbols and verbal descriptions of what the functions do. In our example: f takes the square of a number and adds 1 g doubles a number h subtracts 1 from a number So a verbal description of the composed f@g@h as a sequence of ...

2016年11月25日 · lim_(x rarr 1)(x^2-1)/(x-1) = 2 Let f(x) = (x^2-1)/(x-1) then f(x) is defined everywhere except at x=1, however when we evaluate the limit we are not interested in ...