Asked by lexi
ms sue plss help
Use the function f(x) to answer the questions.
f(x) = −16x2 + 24x + 16
Part A: What are the x-intercepts of the graph of f(x)? Show your work. (2 points)
Part B: Is the vertex of the graph of f(x) going to be a maximum or a 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)
Use the function f(x) to answer the questions.
f(x) = −16x2 + 24x + 16
Part A: What are the x-intercepts of the graph of f(x)? Show your work. (2 points)
Part B: Is the vertex of the graph of f(x) going to be a maximum or a 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)
Answers
Answered by
mathhelper
f(x) = −16x2 + 24x + 16
is a downwards opening parabola, so it must have a maximu.
for the x-intercepts, f(x) = 0
−16x2 + 24x + 16 = 0
divide be -8
2x^2 - 3x - 2 = 0
(2x + 1)(x - 2) = 0
x = -1/2 or x = 2
the parabola cuts the x-axis at (-1/2 , 0) and (2, 0)
for y = ax^2 + bx + c , the x of the vertex is -b/(2a)
so for yours, the x of the vertex is -24/-32 = 3/4
then y = -16(3/4)^2 + 24(3/4) + 16 = 25
You could also complete the square to find the vertex ...
y = -16(x^2 -3/2 x + ....) + 16
= -16(x^2 - 3/2 x + 9/16 - 9/16) + 16
= -16( (x - 3/4)^2 - 9/16) + 16
= -16(x - 3/4)^2 + 9 + 16
= -16(x - 3/4)^2 + 25 , and got the same result
leave it up to you to put it all together
(btw, Ms Sue passed away several years ago, we all miss her)
is a downwards opening parabola, so it must have a maximu.
for the x-intercepts, f(x) = 0
−16x2 + 24x + 16 = 0
divide be -8
2x^2 - 3x - 2 = 0
(2x + 1)(x - 2) = 0
x = -1/2 or x = 2
the parabola cuts the x-axis at (-1/2 , 0) and (2, 0)
for y = ax^2 + bx + c , the x of the vertex is -b/(2a)
so for yours, the x of the vertex is -24/-32 = 3/4
then y = -16(3/4)^2 + 24(3/4) + 16 = 25
You could also complete the square to find the vertex ...
y = -16(x^2 -3/2 x + ....) + 16
= -16(x^2 - 3/2 x + 9/16 - 9/16) + 16
= -16( (x - 3/4)^2 - 9/16) + 16
= -16(x - 3/4)^2 + 9 + 16
= -16(x - 3/4)^2 + 25 , and got the same result
leave it up to you to put it all together
(btw, Ms Sue passed away several years ago, we all miss her)
Answered by
lexi
ohh noo god rest her soul p.s tysm for the help its rlly appreciated