How do i find the tangent to the curve y=5x^3 + 2x -1, where x=1?

Question

Yes its for calculus

Answer 1

f(1) = 5x^3 + 2x -1  = 5 + 2 -1 = 6

so that tangent point is (1, 6)

f(x) = y=5x^3 + 2x -1 ... yes, first just take the derivative

f '(x) = dy/dx = 15x^2 + 2

evaluate when x = 1

f '(1) = 15*1^2 + 2 = 15 + 2 = 17

17 is the slope of

y = mx + b ....

y = 17x + b

to find b, use the tangent point (1, 6)

6 = 17 *1 + b

b = -11

sooooooo

tangent line =

y = 17x -11  <===========  answer

Answer 2

Find y ' first,y'=15x^2 +2,then find its value for x=1,15(1)^2 + 2=17

then the angle of this tan line has Arc tan 17=86.63 degrees with X axis.

Answer 3

f ` (x) = 15 x² + 2

f `(1) = 17 = m

f (1) = 6

Point (1 , 6)

m = 17

y - 6 = 17( x - 1)

y = 17x - 11

Answer 4

If this is calculus you take the derivative.

dy/dx = 15x^2 + 2 = 15 + 2 = 17 at x = 1

This gives you the slope of the tangent so:

y = mx + b = 17x + b

Now x = 1 gives y = 6

And you can use these x and y values to determine b.

Is this for calculus?

Answer 5

dy/dx = 15x^2 + 2

When x = 1

dy/dx = 15(1)^2 + 2

          = 17

Therefore, the gradient of the tangent is 17

When x = 1, y = 6 (1,6) This is where the tangent touches the curve.

Equation of the tangent

y - 6 = 17(x - 1)

y - 6 = 17x - 17

Therefore the equation of the tangent is

y = 17x - 11

Answer 6

OK, the Captain found the slope of the curve at x = 1 for you.

That is also the slope of the line that is the tangent at that point.

The point is where? Substitute x = 1 into the function and y = 5 + 2 - 1 = 6.

So the tangent has slope 17 and goes thru (1,6).

The next answerer can give the the equation for the tangent.

(Who gets the 10 points?)