# Find the equation of the tangent line to x^2+5 at 2?

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The tangent line to a curve, at a given point, represents the derivative of the function of curve, at that point.

We'll determine the 1st derivative of the function:

f'(x) = 2x

But the derivative of a function at a point is the slope of the tangent line at that point:

m = f'(2) = 2*2 = 4

Now, we'll calculate the y coordinate of the point, located on the given curve, whose x coordinate is 2.

f(2) = 4 + 5 = 9

We'll write the equation of a line that passes through a point and it has a slope m.

y - y1 = m(x - x1)

y - 9 = 4(x - 2)

y - 9 = 4x - 8

y = 4x + 1

**The equation of the tangent line to the curve y = x^2 + 5, at the point x = 2, is y = 4x + 1.**