Question

The slope of the tangent line to the curve y = f(x) at (x,y) is y¹ = 10x²+5x. If (1,16) is a point on the curve

Answer 1

$$y'=10x^2+5x$$

$$y=\int(10x^2+5x)dx=\dfrac{10}{3}x^3+\dfrac{5}{2}x^2+C$$

(1,16) is a point on the curve

$$16=\dfrac{10}{3}(1)^3+\dfrac{5}{2}(1)^2+C$$

$$C=\dfrac{61}{6}$$

$$y=\dfrac{10}{3}x^3+\dfrac{5}{2}x^2+\dfrac{61}{6}$$