A tangent line touches a smooth curve at exactly one point, sharing the same instantaneous slope as the curve at that location. Determining its equation is a routine calculus task that relies on the function’s derivative.

Step 1 – Differentiate the Function

Compute f ′(x) using standard differentiation rules. For power functions, f(x)=xⁿ, the power rule gives f ′(x)=n xⁿ⁻¹. For example, for f(x)=2x²+4x+10, the derivative is f ′(x)=4x+4=4(x+1).

When the function is a product, apply the product rule: (f₁f₂)′ = f₁f₂′ + f₁′f₂. For instance, f(x)=x²(x²+2x) yields f ′(x)=x²(2x+2)+2x(x²+2x)=4x³+6x².

Step 2 – Evaluate the Slope at the Desired Point

The slope of the tangent equals the derivative evaluated at the chosen x‑value. For f(x)=2x²+4x+10 at x=5, the slope is m = f ′(5) = 4(5+1) = 24.

Step 3 – Construct the Tangent Line Equation

First find the point of tangency by plugging the x‑value into the original function: f(5)=2·5²+4·5+10=80. Thus the point is (5, 80). Using the point‑slope form y−y₀=m(x−x₀) gives

y−80 = 24(x−5). Rearranging to slope‑intercept form yields y = 24x − 1915.

That final expression is the equation of the tangent line to f(x) at x=5.