are you able to find g '' (x) ?
yes i got \[(x^2 -4x + 16)/((x-2)^2)\]
yep yep did that got plus or minus 4 for g'(x)
If you complete the square for the numerator, you get this x^2 - 4x + 16 x^2 - 4x + 4 - 4 + 16 (x^2 - 4x + 4) + 12 (x-2)^2 + 12 So this means that x^2 - 4x + 16 is always positive. The same applies to (x-2)^2. Overall, g '' (x) is always positive
since g '' (x) is never negative, it's impossible for g to be concave down
okay so if g''(x) is positive then g(x) is concave up?
yeah g is concave up over all of g's domain
okay and as far as i can get on part b is getting the slope ... wich is -7
so now i need a point on the tangent line right?
g(3) = 4 which means (3,4) lies on the g(x) function use m = -7 x = 3 y = 4 with y = mx+b and solve for b
4 = -7(3) + b 25 = b
so y = -7x + 25?
okay great thank you very much !!!
glad to be of help