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# PLZ HELP ALEBRA 2 HW DUE

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When the graph of y=2x^2 -x+7 four units to the right, we obtain the graph of y=ax^2+bx+c. Find a+b+c.

The function f(x) is invertible, but the function g(x)=f(kx) is not invertible. Find the sum of all possible values of k

Given that (-2,3)  is on the graph of y=f(x) , find a point that must be on the graph of y=f(2x+1)+3. Express your answer as an ordered pair (a,b) where a and b are real numbers.

When the graph of a certain function f(x) is shifted 2 units to the right and stretched vertically by a factor of 2 relative to the x-axis (meaning that all y-coordinates are doubled), the resulting graph is identical to the original graph.

Given that f(0)=1 , what is f(10)?

THanks

Sep 28, 2019

#1
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When the graph of y=2x^2 -x+7 four units to the right, we obtain the graph of y=ax^2+bx+c. Find a+b+c.

2x^2 - x + 7

The x coordinate of the vertex is     1 / [2*2]  = 1/4

And the y coordinate of the vertex   =   2(1/4)^2 - (1/4) + 7  =  1/8 + 1/4 + 7  =    55/8

When shifted four units to the right the vertex becomes   (17/4, 55/8)

So we have that

2(x -17/4)^2 + 55/8       simplify

2(x^2 - (17/2)x + 289/16)  + 55/8

2x^2 - 17x + 289/8 + 55/8

2x^2 - 17x + 43

So

a + b + c  =   2  - 17   +  43   =    28   Sep 28, 2019
#2
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A different approach:

2(x-4)^2 -(x-4) + 7 =

2(x^2-8x+16) -x+4+7 =

2x^2 -17x+ 43

a+b+c = 2-17+43

Sep 28, 2019
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