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.

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Given that f(0)=1 , what is f(10)?

THanks

Guest Sep 28, 2019

#1**+4 **

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

CPhill Sep 28, 2019

#2**+4 **

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

ElectricPavlov Sep 28, 2019