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HelloKitty17
 one year ago
An expression is shown below:
f(x) = 16x2 + 8x + 3
Part A: What are the xintercepts of the graph of the f(x)? Show your work. (2 points)
Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points)
Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points)
HelloKitty17
 one year ago
An expression is shown below: f(x) = 16x2 + 8x + 3 Part A: What are the xintercepts of the graph of the f(x)? Show your work. (2 points) Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points) Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points)

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OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.1A) To get xintercepts, set f(x) (or y) to zero and solve for x. Solve 16x^2 + 24x + 16 = 0 divide throughout by 8 2x^2  3x  2 = 0 solve for x. 2x^2  3x  2 = 0 2x^2  4x + x  2 = 0 2x(x  2) + 1(x  2) = 0 (2x+1)(x2) = 0 x = 0.5 or x = 2 The xintercepts are: 0.5 and 2. B) Is the vertex of the graph of f(x) going to be a maximum or minimum? f(x) = 16x^2 + 24x + 16 The "a" or the coefficient of x^2 is 16. Since it is a negative, it is an inverted parabola that opens downward. Therefore, the vertex will be a maximum. To find the vertex, complete the square of 16x^2 + 24x + 16 The vertex form of this equation will look like: f(x) = a(xh)^2 + k where (h,k) is the vertex. Complete the square of 16x^2 + 24x + 16 Make the coefficient of x^2 1 by factoring out 16 16(x^2  24/16x  1) = 16(x^2  3/2x  1). To complete the square: Divide the coefficient of x by 2: 3/2 / 2 = 3/4. This will go inside the parenthesis to be squared and you will have to subtract the square of it: 16(x^2  3/2x  1) = 16{ (x3/4)^2  (3/4)^2  1 } = 16{ (x3/4)^2  9/16  1 } = 16{ (x3/4)^2  25/16 } = f(x) = 16(x3/4)^2 + 25 compare this to f(x) = a(xh)^2 + k where (h,k) is the vertex and you can see that h = 3/4 and k = 25. So the vertex is (3/4, 25) or (0.75, 25) C) f(x) = 16x^2 + 24x + 16 We know the graph is a parabola. We also know it is an inverted parabola because of the negative coefficient of x^2. We know the curve crosses the xaxis at x = 0.5 and x = 2 as they are the xintercepts found in part A). From Part B) we know that the vertex of the parabola is at (0.75, 25). We can also set x = 0 and find the yintercept: 16(0)^2 + 24(0) + 16 = 16. So the curve crosses the yaxis at y = 16 We have enough information to plot the graph.
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