Asked by girly girl
1. What is the equation of the line tangent to the function f(x)=-x^2+5 at the point (1,4)?
a. y=-x+5
b. y=-x+9
c. y=-2x+6
d. y=-2x+9
2. Given f(x)=4x^2+2x-9, what is f'(a)?
a. f'(a)=6a+2
b. f'(a)=6a-9
c. f'(a)=8a+2
d. f'(a)=4a+2-9
a. y=-x+5
b. y=-x+9
c. y=-2x+6
d. y=-2x+9
2. Given f(x)=4x^2+2x-9, what is f'(a)?
a. f'(a)=6a+2
b. f'(a)=6a-9
c. f'(a)=8a+2
d. f'(a)=4a+2-9
Answers
Answered by
Damon
f'(x) = -2x
at x = 1
f'(x) = slope m = -2
so
y = -2x + b
at (1,4)
4 = -2 + b
b = 6
continue
f' = 8 x + 2
so put in a for x
what is your problem with this?
at x = 1
f'(x) = slope m = -2
so
y = -2x + b
at (1,4)
4 = -2 + b
b = 6
continue
f' = 8 x + 2
so put in a for x
what is your problem with this?
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