 # For the function, find the point(s) on the graph at which the tangent line has slope 3. y=1/3 x^3-3x^2 +11x+15 Sandra Terrell 2022-08-08 Answered
For the function, find the point(s) on the graph at which the tangent line has slope 3.
$y=\frac{1}{3}{x}^{3}-3{x}^{2}+11x+15$
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$y=\frac{{x}^{3}}{3}-3{x}^{2}+11x+15$
$⇒\frac{dy}{dx}={x}^{2}-6x+11$
Given, slope of tangent, i.e., $\frac{dy}{dx}=3$
$\therefore {x}^{2}-6x+11=3$
$⇒{x}^{2}-6x+8=0$
$⇒\left(x-4\right)\left(x-2\right)=0$
$⇒x=2,4$
Thus, corresponding values of y are:
(i) for x = 2
$y=\frac{{2}^{3}}{3}-3×{2}^{2}+11×2+15=\frac{83}{3}$
(ii) for x = 4
$y=\frac{97}{3}$
So, required points are (2,83/3) and (4,97/3).
###### Not exactly what you’re looking for? sublimnes9
${y}^{\prime }={x}^{2}-6x+11$
let slop $={y}^{\prime }={x}^{2}-6x+11=3$
$⇒{x}^{2}-6x+8=0$
$⇒\left(x-2\right)\left(x-4\right)=0$
x=2, 4
when x=2
$y=\left(1/3\right){2}^{3}-3\ast {2}^{2}+11\ast 2+15=83/3$
x=4
$y=1/3\ast {4}^{3}-3\ast {4}^{2}+11\ast 4+15=97/3$
so the points which the slop is 3 are (2,83/3) and(4,97/3)