Answer:oyoyoyohkyhiibjgi
Step-by-step explanation:
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What do you need help on sir
Answer:
Real values of x where x < -1
Step-by-step explanation:
Above the x-axis, the function is positive.
The function is decreasing when the gradient is negative.
The function has a positive

coefficient, therefore the vertex is a local minimum;
This means the gradients are negative before the vertex and positive after it;
To meet the conditions therefore, the function must be before the vertex and above the x-axis;
This will be anywhere before the x-intercept at x = -1;
Hence it is when x < -1.
He first two are relatively easy. f(x-2) shifts it sideways (2 to the right in this case) f(x) -2 shifts it straight down (by 2)
f(2x) means replace x with 2x in the original 4(2x + 1)^2 − 3 if you factor out the 2 from the square root you get 4*sqr(2) ( x + ½)^2 - 3 which is now in the form a ( x -h)^2 + k the vertex is now at (-½, -3) (it used to be at (-1,-3) the "a" has gotten bigger, which makes the parabola "skinnier" (it goes up faster)
2•f(x) becomes 2*( 4(x + 1)2 − 3 ) = 8 (x+1)^2 -6 where is the vertex now? is this parabola fatter or skinnier than the original f(x) ?