SOLUTION: please help with this math question from blitzer 4 (1) use the graph of y =x^2 -2x -8, what are the solutions to this equation? (b) does this unction have a maximum or a mini

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Question 135309: please help with this math question from blitzer 4
(1) use the graph of y =x^2 -2x -8, what are the solutions to this equation?
(b) does this unction have a maximum or a minimum ?
(c) what is the equation of the line of symmetry for this graph?
(d) what are the coordinates of the vertex in (x,y) form?

(2) calculate the value of the discriminant of x^2 + 2x +1 =0
(b)how many x intercepts would the graph of y =x^2 +2x +1
(3) if a rock is throw upward with an initial velocity of 40 ft per second from the top of a 30 ft building, write the height as so, velocity as vo and t as time. equation being s=-16t^2+40+30
(a) how high is the rock after 2 seconds?
(b) how many second will the graph reach maximum height?
(c) what is the maximum height?
i know it's a lot ,but i really need your help. thank you so much

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(1) use the graph of y =x^2 -2x -8, what are the solutions to this equation?
graph%28400%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E2-2x-8%29
Solutions: x = 4 or x = -2
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(b) does this function have a maximum or a minimum ?
It has a minimum because the coefficient of x^2 is positive.
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(c) what is the equation of the line of symmetry for this graph?
x = -b/2a = -(-2)/2 = 1
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(d) what are the coordinates of the vertex in (x,y) form?
Vertex: (1,f(1) which is (1,-9)
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(2) calculate the value of the discriminant of x^2 + 2x +1 =0
b^2-4ac = 2^2-4*1*1 = 0
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(b)how many x intercepts would the graph of y =x^2 +2x +1
It meets the x-axis at x=-1 but does not pass thru it.
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(3) if a rock is throw upward with an initial velocity of 40 ft per second from the top of a 30 ft building, write the height as so, velocity as vo and t as time. equation being s=-16t^2+40t+30
(a) how high is the rock after 2 seconds?
s(2) = -16(2)^2+40(2)+30 = -64+80+30 = 46 ft
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(b) how many second will the graph reach maximum height?
t = -b/2a = -40/(2*-16) = 1.25 seconds
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(c) what is the maximum height?
s(1.25) = -16(1.25)^2+40*1.25+30 = 55 ft.
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Cheers,
Stan H.