Drag the values to the correct locations on the image. Not all values will be used.Function f is a logarithmic function with a vertical asymptote at 𝑥=0 and an x-intercept at (4,0). The function is decreasing over the interval (0,∞).Function g is represented by the equation 𝑔⁡(𝑥)=log2⁡(𝑥+3)−2.Over which interval are both functions positive?014-∞∞2

Use the drawing tools to form the correct answers on the graph.Draw the lines representing the vertical and horizontal asymptotes on the graph of this function.𝑓⁡(𝑥)=2⁢𝑥2+7⁢𝑥−4𝑥2+5⁢𝑥+4

Sketch the graph of a function that satisfies all of the given conditions.vertical asymptote x = 0,f ′(x) > 0 if x < −3,f ′(x) < 0 if x > −3 (x ≠ 0),f ″(x) < 0 if x < 0,f ″(x) > 0 if x > 0

The graph of a function f is given in the figure.A curve is shown on the x y coordinate plane. It begins at the point (−2, −1), goes up and to the right, passes through the approximate point (−1, −0.2), and passes through the negative x-axis at the approximate point (−0.8, 0). It then continues up and right, passes through the positive y-axis at the point (0, 1), and reaches a high point at (1, 3). It then goes down and right, passes through the points (2, 2) and (3, 1), and ends at the approximate point (4, 0.5).(a)Find the value of f(1).(b)Estimate the value of f(−1).(c)For what values of x is f(x) = 1? (Enter your answers as a comma-separated list.) (d)Estimate the value of x such that f(x) = 0.x = (e)State the domain and range of f. (Enter your answers in interval notation.)domain range (f)On what interval is f increasing? (Enter your answer using interval notation.)

To determine the image set (range) of the function \( f(x) = (x - 4)^2 + 4 \) for the domain \(-2 < x < 4\), we need to analyze the behavior of the function within the given interval. First, let's rewrite the function: \[ f(x) = (x - 4)^2 + 4 \] The function \( (x - 4)^2 \) is a parabola that opens upwards with its vertex at \( x = 4 \). The minimum value of \( (x - 4)^2 \) is 0, which occurs when \( x = 4 \). Therefore, the minimum value of \( f(x) \) is: \[ f(4) = (4 - 4)^2 + 4 = 0 + 4 = 4 \] Next, we need to find the maximum value of \( f(x) \) within the interval \(-2 < x < 4\). Since \( (x - 4)^2 \) increases as \( x \) moves away from 4, the maximum value of \( f(x) \) will occur at the left endpoint of the interval, \( x = -2 \): \[ f(-2) = (-2 - 4)^2 + 4 = (-6)^2 + 4 = 36 + 4 = 40 \] Therefore, the image set (range) of the function \( f(x) = (x - 4)^2 + 4 \) for \(-2 < x < 4\) is: \[ 4 < f(x) < 40 \] So, the correct answer is: **D. (4, 40)**

The graph of a function g is shown.The x y-coordinate plane is given. A function labeled y = g(x) with 4 parts is graphed.The first part is a curve, enters the window in the second quadrant, goes up and right becoming less steep, crosses the y-axis at approximately y = 2.5, and ends at the open point (2, 3).The second part is a curve begins again at the open point (2, 1), goes up and right becoming less steep, and ends at the open point (5, 2).The third part is the closed approximate point (5, 1.2).The fourth part is a curve, begins at the open point (5, 2) goes down and right becoming more steep, and exits the window in the first quadrant.Use it to state the values (if they exist) of the following:(a)  lim x → 2− g(x)(b)  lim x → 2+ g(x)(c)  lim x → 2 g(x)(d)  lim x → 5− g(x)(e)  lim x → 5+ g(x)(f)  lim x → 5 g(x)SolutionLooking at the graph we see that the values of g(x) approach as x approaches 2 from the left, but they approach as x approaches 2 from the right.Therefore (a) lim x → 2− g(x) =     and    (b) lim x → 2+ g(x) = .Since the left and right limits are different, we conclude that (c) the limit as x approaches 2 of g(x) does not exist.The graph also shows that (d) lim x → 5− g(x) =     and    (e) lim x → 5+ g(x) = .This time, the left and right limits are the same and so, by this theorem, we have (f) lim x → 5 g(x) = Despite this fact, notice that g(5) ≠ 2.