Question

Find the equation of the line tangent to the function at the given point f(x)=5x^2 at x=10

Answer 1

f(x) = 5x²

f(10) = 5(10)² = 5(100) = 500

f'(x) = 10x

f'(10) = 10(10) = 100

Now, find the line that passes through (10, 500) and has a slope of 100

y - y₁ = m(x - x₁)

y - 500 = 100(x - 10)

y - 500 = 100x - 1000

y = 100x - 500