Question

f(x)=x^4-25x+144, x=1 Find the slope of the tangent line to graph the given value of x

Answer 1

you need to find f'(1), or the derivative of f(x) evaluated at 1.
f'(x) = 4x^3 - 25
f'(1) = 4-25
= -21
the slope of the tangent line (which is the slope of the line) at x=1 is -21.

Answer 2

im soo sorry thats thhe wrong problem its : f(x)=x^4-25x^2+144, x=1 Find the slope of the tangent line to graph the given value of x

Answer 3

f'(x) = 4x^3 - 50x
f'(1) = 4 - 50
=-46

Answer 4

is -46 y or is it the slope?

Answer 5

its the slope

Answer 6

ok

Answer 7

but where y?

Answer 8

umm, so i gave you the slope to the tangent line, would you like the entire equation of the tangent line? at x = 1, y is...
f(1) = 4 - 25 + 144
= 123
the tangent line is of the form y = m*x + b, where m equals the slope, -46. now plug in y = 123 at x = 1 to get b...
123 = -46*1 + b
b = 169

the equation of the tangent line is y = -46*x + 169
and y at x = 1 is 123 if you wanted that too

Answer 9

tht should be 120 shouldnt it be?

Answer 10

(1)^4-25(1)^2+144=120

Answer 11

oh yeah, sorry

Answer 12

its cool