$y=\frac{1}{3}{x}^{3}-3{x}^{2}+11x+15$

Jazmin Clark
2022-08-08
Answered

For the function, find the point(s) on the graph at which the tangent line has slope 3.

$y=\frac{1}{3}{x}^{3}-3{x}^{2}+11x+15$

$y=\frac{1}{3}{x}^{3}-3{x}^{2}+11x+15$

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pokajalaq1

Answered 2022-08-09
Author has **18** answers

$y=\frac{{x}^{3}}{3}-3{x}^{2}+11x+15$

$\Rightarrow \frac{dy}{dx}={x}^{2}-6x+11$

Given, slope of tangent, i.e., $\frac{dy}{dx}=3$

$\therefore {x}^{2}-6x+11=3$

$\Rightarrow {x}^{2}-6x+8=0$

$\Rightarrow (x-4)(x-2)=0$

$\Rightarrow x=2,4$

Thus, corresponding values of y are:

(i) for x = 2

$y=\frac{{2}^{3}}{3}-3\times {2}^{2}+11\times 2+15=\frac{83}{3}$

(ii) for x = 4

$y=\frac{97}{3}$

So, required points are (2,83/3) and (4,97/3).

$\Rightarrow \frac{dy}{dx}={x}^{2}-6x+11$

Given, slope of tangent, i.e., $\frac{dy}{dx}=3$

$\therefore {x}^{2}-6x+11=3$

$\Rightarrow {x}^{2}-6x+8=0$

$\Rightarrow (x-4)(x-2)=0$

$\Rightarrow x=2,4$

Thus, corresponding values of y are:

(i) for x = 2

$y=\frac{{2}^{3}}{3}-3\times {2}^{2}+11\times 2+15=\frac{83}{3}$

(ii) for x = 4

$y=\frac{97}{3}$

So, required points are (2,83/3) and (4,97/3).

Sandra Terrell

Answered 2022-08-10
Author has **2** answers

${y}^{\prime}={x}^{2}-6x+11$

let slop $={y}^{\prime}={x}^{2}-6x+11=3$

$\Rightarrow {x}^{2}-6x+8=0$

$\Rightarrow (x-2)(x-4)=0$

x=2, 4

when x=2

$y=(1/3){2}^{3}-3\ast {2}^{2}+11\ast 2+15=83/3$

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)

let slop $={y}^{\prime}={x}^{2}-6x+11=3$

$\Rightarrow {x}^{2}-6x+8=0$

$\Rightarrow (x-2)(x-4)=0$

x=2, 4

when x=2

$y=(1/3){2}^{3}-3\ast {2}^{2}+11\ast 2+15=83/3$

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)

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