Exam 3: Quadratic Functions and Equations; Inequalities

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Consider only the discriminant, b24acb ^ { 2 } - 4 a c to determine whether one real-number solution, two different real-number solutions, or two different imaginary-number solutions exist. - 7x2=21x7 x ^ { 2 } = 21 x

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Solve by completing the square to obtain exact solutions. - 25x2+50x+16=025 x ^ { 2 } + 50 x + 16 = 0

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Solve graphically. Round solutions to three decimal places, where appropriate. - 3x25=6x3 x ^ { 2 } - 5 = 6 x

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Determine whether there is a maximum or minimum value for the given function, and find that value. - f(x)=x2+5x3f ( x ) = - x ^ { 2 } + 5 x - 3

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Solve the problem. -Your company uses the quadratic model y=11x2+350xy = - 11 x ^ { 2 } + 350 x to represent how many units ( yy ) of a new product will be sold (x)( x ) weeks after its release. How many units can you expect to sell in week 12?

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Solve. - 4x+73=3\sqrt [ 3 ] { 4 x + 7 } = - 3

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Solve. - 8y+52y5=10y225\frac { 8 } { y + 5 } - \frac { 2 } { y - 5 } = \frac { 10 } { y ^ { 2 } - 25 }

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Determine whether there is a maximum or minimum value for the given function, and find that value. - f(x)=12x22x152f ( x ) = \frac { 1 } { 2 } x ^ { 2 } - 2 x - \frac { 15 } { 2 }

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Use the given graph to find the x-intercepts and zeros of the function. -Use the given graph to find the x-intercepts and zeros of the function. -

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Find the axis of symmetry of the given function. - f(x)=x217x+10f ( x ) = x ^ { 2 } - 17 x + 10

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Graph. -Graph. -

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Match the equation to the correct graph. - y=x27y = x ^ { 2 } - 7

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Use the quadratic formula to find the exact solutions. - x2=15+3xx ^ { 2 } = 15 + 3 x

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Find the vertex of the parabola. - g(x)=x213x+6g ( x ) = x ^ { 2 } - 13 x + 6

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Match the equation to the correct graph. - y=(x+3)2+2y = - ( x + 3 ) ^ { 2 } + 2

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Graph. - f(x)=4x23x8f(x)=4 x^{2}-3 x-8  Graph. - f(x)=4 x^{2}-3 x-8

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Solve the inequality and graph the solution set. - x>5| x | > 5  Solve the inequality and graph the solution set. - | x | > 5

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Solve. - 12x330x2+10x25=012 x ^ { 3 } - 30 x ^ { 2 } + 10 x - 25 = 0

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Find the vertex of the parabola. - f(x)=2x212x+16f ( x ) = 2 x ^ { 2 } - 12 x + 16

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Consider only the discriminant, b24acb ^ { 2 } - 4 a c to determine whether one real-number solution, two different real-number solutions, or two different imaginary-number solutions exist. - 25x210x+1=025 x ^ { 2 } - 10 x + 1 = 0

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