Question

Find an equation of the tangent line to the curve at the given point.
(4x^4)+(4x^2)-8x with point of (1,0)

Answer 1

\[4x ^{4}+4x ^{2}-8x\]

Answer 2

y'=16x^3+8x-8=8(2x^3+x-1)
y' at x=1 is 8(2+1-1)=16
y' represents the slope
so the slope is 16
y=mx+b
y=16x+b
we know a point on the tangent line (1,0)
0=16(1)+b
so we can solve for b
so b=-16
so the tangent line
is y=16x-16

Answer 3

Why you use x =1? when you solved to get 16

Answer 4

we want to find the tangent line at x=1
so we need to find the slope at x=1

Answer 5

so it it possinle yo use any other x value?

Answer 6

Or becasue the opint given is (1,0)

Answer 7

it depends at what point they want the tangent line
they asked for the tangent line at x=1 so we found the tangent line at x=1

Answer 8

ok well now i get thanks again

Answer 9

(x=1,y=0)