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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to write the equation of the tangent line to the given curve, at the point `x = 1` , such that:

`f'(1) = (f(x) - f(1))/(x - 1)`

You need to evaluate derivative of the function at `x = 1` , such that:

`f'(x) = 10x + 0 => f'(1) = 10*1 => f'(1) = 10`

You need to evaluate `f(x)` at `x = 1` , such that:

`f(1) = 5*1^2 + 1 => f(1) = 6`

Substituting `10` for `f'(1)` and `6` for `f(1)` in equation of tangent line, yields:

`10 = (f(x) - 6)/(x - 1) => 10(x - 1) = (f(x) - 6)`

`f(x) = 10x - 10 + 6 => f(x) = 10x - 4`

Hence, evaluating the equation of tangent line to the given curve, at `x = 1` , yields `f(x) = 10x - 4.`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The derivative of a function at a point is the slope of the tangent line, at the point.

The equation of the tangent line, at 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

We'll differentiate the given function with respect to x:

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

f'(x) = 10x

Now, we'll replace 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

We'll add 6 both sides to isolate y to the left:

y = 10x - 10 + 6

y = 10x - 4

The equation of the tangent line, to the curve y = 5x^2 + 1, is: y = 10x - 4.

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