## Calculus: Early Transcendentals 8th Edition

$f(1) = 8$
We know that: $f'(x)=3-4x$ Thus: $f(x) = 3x - 2x^2 + C$ Use the information given of $(2,5)$: $5 = 3(2) - 2(2)^2 + C$ $5 = 6 - 8 + C$ $5 = -2 + C$ $7 = C$ $f(x) = 3x - 2x^2 + 7$ Substitute $x = 1$ $f(1) = 3(1) - 2(1)^2 + 7$ $f(1) = 3 - 2 + 7$ $f(1) = 8$