Question 1173487
f(x)=2x^4+ax^3-4x^2+bx-18
f(2)=4=32+8a-16+2b-18
so 6=8a+2b
f(1)=0=2+a-4+b-18
so 20=a+b
a=20-b
6=160-8b+2b
-154=-6b
b=25 2/3
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b=20-a
6=8a+40-2a
-34=6a
a=-5 2/3
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The function is f(x)=2x^4-(17/3)x^3-4x^2+(77/3)x-18
When x=1 f'(x)=8x^3-17x^2-8x+(77/3) and f(1)=8-17-8+77/3=8 2/3 units, instantaneous rate of change.
f(1)=0 so equation of tangent line at (1, 0) is y=(26/3)x-26/3
set derivative equal to 0.
8x^3-17x^2-8x+(77/3)-18=0
try 0 and slope is 7 2/3 so no turning point
at  x=-1 slope is still positive
but at x=-2 the slope is strongly negative
estimate the turning point (a local minimum) between x=-1 and x=-2 (it is at x=-1.19)
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{{{graph(300,300,-10,10,-55,50,2x^4-(17/3)x^3-4x^2+(77/3)x-18,(26x/3)-(26/3))}}}