# A line is tangent to the curve y=5x^2+1, in a point x=1. What is the equation of tangent line.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

If y = f(x), then the tangent of the curve at x = x1 is given by:

y -y1 = f'(x1)(x-x1),.............(1)where y1 = f(x1) .

The given curve is y = f(x) =5x^2+1.

Given x1 = 1

Therefore y1 = f(1) = 5*1^2+1 = 5.

y' = (5x^2+1)'

y' = 2*5x

at x = 1, y' = 2*5*1 = 10.

y1 = f(1)

Therefore the  equation of the tangent at x = 1 is got  by substituting the values of x1 = 1, y1 = 6 and f'(x1) = f'(1) = 10 in y -y1 = f'(x1)(x-x1).

y-6 = 10(x-1). Or

10x-y-10+6 = 0

10x-y-4 = 0

So 10x-y -4 is the required tangent.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The tangent line in a point is the derivative of a function in that point.

The equation of the tangent line, in the point x = 1 is:

y - f(1) = f'(1)(x-1)

We'll calculate f(1), by substituting x by 1 in the expression of the function:

f(1) = 5*1^2 + 1

f(1) = 5 + 1

f(1) = 6

To calculate f'(1), first we'll have to differentiate the given function:

f'(x) = (5x^2 + 1)'

f'(x) = 10x

Now, we'll substitute x by 1 in the expression of the first derivative:

f'(1) = 10

Now, we'll substitute f(1) and f'(1) in the expression of the equation of the tangent line:

y - f(1) = f'(1)(x-1)

y - 6 = 10(x - 1)

We'll remove the brackets:

y - 6 = 10x - 10