Question

find a and b if the equation of the tangent line to the graph of the function f(x)=x^(4)+ax^(2)+b at x=1 is y=-4x+3 this still has me pretty confused

Answer 1

f'(x)=4x^3+2ax
f'(1)=-4=>4+2a=-4=>a=-4
now x=1=>y=-4x+3=-1
substitute in f(x)=>-1=1-4+b=>b=2

Answer 2

kind of lost me there.

Answer 3

the slope of y = -4x + 3 is -4
so, -4 = f'(1) = 4 + 2a
-4 = 4 + 2a
-4 -4 = 2a
-8 = 2a
-4 = a

no from original equaation
F(x) = x^(4)+ax^(2)+b
-1 = 1^4 + -4(1)^2 + b
-1 = 1 + -4 + b
-1 - 1 + 4 = b
b = 2

Answer 4

the derivative of a function at any point gives the slope of the tangent line
the tangent line will be of the form:
y = mx +b
m = slope = f'(x)
y = f(x)