please help by answering and showing work!

Question

1. Use the vertex formula to find the vertex. Find all intercepts and then sketch. Dot in the line of symmetry.

f(x) = x2 – 6x + 5

Vertex: ( , ) y-intercept: ( , )

x-intercepts: ( , ) , ( , )

equation of the line of symmetry: ______________

2. Consider the quadratic model h(t) = -16 2 t + 40t + 50 for the height (in feet), h, of an object t seconds after the object has been projected straight up into the air. a) Find the maximum height attained by the object. Use the vertex formula. b) How much time does it take to fall back to the ground?

3. You have 100 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. HINT: Use perimeter and area concepts to relate L and W, then write the area of the rectangle as a function of just L or W. It will be quadratic function, with maximum at its vertex.

L

W

4. . Given f(x) = −1/2(𝑥 + 5)^3(𝑥 + 1)(𝑥 − 2)^2 , list the zeros and their multiplicities. What is the degree of the polynomial? What is the end behavior model? What is the y-intercept? Make a rough sketch using this information. You can check using the graphing calculator, but keep in mind that the grapher may mislead you if you don’t adjust the window!

Zeros and their multiplicities: _________________________________

degree of f(x) : ________________________

end behavior model : What power function y = a 𝑥𝑛 does it resemble for very large or very small values of x? ____________________

y-intercept: ( , )

Sketch of f(x):

5. Graphing calculator needed. Be sure to label any graphs with the “window” you used. Use “Calc” and “Stat Calc” for more precise approximations. Round answers to the nearest hundredths. Show your work where appropriate. Write the equations for each.

Enter the data below on your grapher and sketch the scatter plot it shows. Use Stat Calc to find both the best fit linear and then the best fit quadratic functions. Sketch both of those on your graph below.

X 0 1 2 3 4 5 6

Y 1.2 1 2 3.5 5.8 12 17

f(x) = ____________x + _____________

f(x) = ___________x2 + ____________x + ____________

Which model do you think fits the data better and why?

Answer 1

(1)

symmetry line is -b/(2a) = -(-6)/(2*1) = 6/2 = 3

f(3) = 3^2 - 6(3) + 5 =

9 - 18 + 5 =

-9 + 5

-4

vertex at (3,-4)

f(x) = (x -5)(x-1), so the solutions are x=5,1;

note that the symmetry axis is at the AVERAGE of these

solutions which is (5+1)/2 = 6/2 = 3

the x-intercepts are (5,0) and (1,0)

the y-intercept is (0,5)

you can graph at desmos dot com

(2) a=-16 b=40 c = 50

-b/(2a) = -40/(2*-16) = -40/-32 = 5/4

h(5/4) = -16(5/4)^2 + 40(5/4) + 50 =

-16( 25/16) + 40(5/4) + 50 =

-25 + 50 + 50 = 75

max at t=5/4 seconds, max height is 75

hits the ground when height function is zero.... solves the quadratic equation

quadratic formula says

[-40 +or- sqrt( 1600 + 4*16*50)]/-32

[-40 +or- sqrt( 1600 + 3200)]/ -32

[ -40 +or- sqrt(4800)]/-32

[ -40 +or- sqrt(16*3*100)]/ -32

[ -40 +or- 4*10*sqrt(3)]/ -32

[ -40 +or- 40*sqrt(3)] / -32

40 [-1 +or- sqrt(3)] / -32

5[-1 +or- sqrt(3)]/-4

(-5/4) (-1 +or- sqrt(3))

positive branch leads to negative measures

(-5/4)( -1 - sqrt(3)) =

(5/4)(1 + sqrt(3)) which is approxiamtely 3.415

(3) perimeter is 100 ---> 2w+L = 100 ---> L = 100 - 2w

Area function A(w) = w ( 100 - 2w) = 100w - 2w^2

max occurs at w= -100/(2*-2) = -100/-4 =25

width w=25 and length = 100 - 2*25 = 50 ---> max area is 1250

(4) degree 6

solutions x=-5 with multiplicity 3

x = -1

x = 2 with multiplicity 2

y intercept : f(0) = (-1/2)(125)(4) = -250

(5)

ŷ = 2.61429X - 1.77143

correlation coefficient r 0.9961944525

C 1.41904762

B -1.214285714

A 0.638095238

y = Ax^2 + Bx+ C

did not get the correlation coefficient R for the linear regression model, but

the quadratic one here is going to be tough to beat

My money is on the parabola