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
+128475 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
+36916 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
