[-/ $3$ Points] $\newline$ DETAILS $\newline$ MY NOTES $\newline$ LARCALCPRECA $\newline$ Find $k$ such that the line is tangent to the graph of the function. $\newline$ \begin{tabular}{|l|c|} $\newline$ \hline Function & Une \\ $\newline$ \hline $f(x)=k x^{3}$ & $y=5 x+9$ \\ $\newline$ \hline $\newline$ \end{tabular} $\newline$ $k=$ $\newline$ $\square$ $\newline$ Need Help? $\newline$ Read it

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Algebra 2

Solve a quadratic equation using the zero product property

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Q. [-/

3

Points]

\newline

DETAILS

\newline

MY NOTES

\newline

LARCALCPRECA

\newline

Find

k

such that the line is tangent to the graph of the function.

\newline

\begin{tabular}{|l|c|}

\newline

\hline Function & Une \\

\newline

\hline

f(x)=k x^{3}

y=5 x+9

\newline

\hline

\newline

\end{tabular}

\newline

k=

\newline

\square

\newline

Need Help?

\newline

Read it

Find Tangent Line Slope: To find the value of $k$ for which the line $y=5x+9$ is tangent to the graph of $f(x)=kx^3$ , we need to ensure that the line and the curve have the same slope at the point of tangency. The slope of the line is given by the coefficient of $x$ , which is $5$ . Therefore, we need to find the derivative of $f(x)$ and set it equal to $5$ to find the point of tangency.
Derivative of $f(x)$ : First, let's find the derivative of $f(x)=kx^3$ . The derivative $f'(x)$ with respect to $x$ is found using the power rule, which states that the derivative of $x^n$ is $n\cdot x^{(n-1)}$ . $\newline$ $f'(x) = \frac{d}{dx} [kx^3] = 3kx^2$
Set Derivative Equal: Now, we set the derivative equal to the slope of the tangent line, which is $5$ . $\newline$ $3kx^2 = 5$ $\newline$ We need to find the value of $k$ that makes this equation true for some value of $x$ where the line and the curve intersect.
Find Intersection Point: Since we are not given a specific point of tangency, we need to find the $x$ -coordinate where the line and the curve intersect by setting the equations of the line and the curve equal to each other. $kx^3 = 5x + 9$
Calculate Discriminant: To solve for $x$ , we can try to factor the equation or use a numerical method. However, since we are looking for a general $k$ that makes the line tangent to the curve, we can use the fact that at the point of tangency, there will be exactly one solution for $x$ . This means the discriminant of the cubic equation must be zero.
Discriminant Formula: The discriminant of a cubic equation $ax^3 + bx^2 + cx + d = 0$ is given by $\Delta = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$ . However, in our equation $kx^3 - 5x - 9 = 0$ , we have $a = k$ , $b = 0$ , $c = -5$ , and $d = -9$ . Plugging these into the discriminant formula simplifies it significantly since terms involving $b$ will be zero. $\newline$ $\Delta = 18\cdot k\cdot 0\cdot (-5)\cdot (-9) - 4\cdot 0^3\cdot (-9) + 0^2\cdot (-5)^2 - 4\cdot k\cdot (-5)^3 - 27\cdot k^2\cdot 81$ $\newline$ $\Delta = 0 - 0 + 0 - 4\cdot k\cdot (-125) - 27\cdot k^2\cdot 81$ $\newline$ $\Delta = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$ $0$
Discriminant Calculation: For the line to be tangent to the curve, the discriminant $\Delta$ must be zero. $\newline$ $500k - 2187k^2 = 0$ $\newline$ $k(500 - 2187k) = 0$
Solve for k: This equation has two solutions for k: $k = 0$ or $k = \frac{500}{2187}$ . However, $k = 0$ would make the function $f(x) = 0$ , which would not result in a cubic function. Therefore, we discard $k = 0$ and take the second solution. $\newline$ $k = \frac{500}{2187}$
Discard $k=0$ : We simplify the fraction $\frac{500}{2187}$ to its simplest form. $\newline$ $k = \frac{500}{2187}$