Question

find all points (x, y) on the graph of f(x)=$\frac{1}{3}x^{3}-4x^{2}+20x + 23$ with tangent lines parallel to the line 15x - 3y = 2. the point(s) is/are (simplify your answer. type an ordered pair using integers or fractions. use a comma to separate answers as needed )

Explanation:

Step1: Find the slope of the given line

Rewrite $15x - 3y=2$ in slope - intercept form $y = mx + b$.
$3y=15x - 2$, so $y = 5x-\frac{2}{3}$. The slope $m = 5$.

Step2: Find the derivative of the function $f(x)$

$f(x)=\frac{1}{3}x^{3}-4x^{2}+20x + 23$. Using the power rule $(x^n)'=nx^{n - 1}$, $f'(x)=x^{2}-8x + 20$.

Step3: Set the derivative equal to the slope of the line

Set $f'(x)=5$, so $x^{2}-8x + 20 = 5$.
Rearrange to get $x^{2}-8x+15 = 0$.
Factor the quadratic equation: $(x - 3)(x - 5)=0$.
Solve for $x$: $x=3$ or $x = 5$.

Step4: Find the corresponding $y$ - values

When $x = 3$, $y=f(3)=\frac{1}{3}(3)^{3}-4(3)^{2}+20(3)+23=9-36 + 60+23=56$.
When $x = 5$, $y=f(5)=\frac{1}{3}(5)^{3}-4(5)^{2}+20(5)+23=\frac{125}{3}-100 + 100+23=\frac{125}{3}+23=\frac{125 + 69}{3}=\frac{194}{3}$.

Answer:

$(3,56),(5,\frac{194}{3})$