Question #333045

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

1
Expert's answer
2022-04-27T12:23:31-0400
y=10x2+5xy'=10x^2+5x

y=(10x2+5x)dx=103x3+52x2+Cy=\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=103(1)3+52(1)2+C16=\dfrac{10}{3}(1)^3+\dfrac{5}{2}(1)^2+C

C=616C=\dfrac{61}{6}

y=103x3+52x2+616y=\dfrac{10}{3}x^3+\dfrac{5}{2}x^2+\dfrac{61}{6}


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