if the slope of the tangent line to a curve is 13 at a certain point. find the function and its point.\n\n$y…

Answer:

C. $y = 3x^{3}+4x;(-1,0)$

Explanation:

Step1: Recall derivative - slope relationship

The slope of the tangent line to a curve $y = f(x)$ at a point is given by $y'=f'(x)$.

Step2: Differentiate option A

For $y = 2x^{2}+2$, using the power - rule $(x^n)'=nx^{n - 1}$, we have $y'=4x$. At the point $(2,2)$, $y'(2)=4\times2 = 8\neq13$.

Step3: Differentiate option B

For $y = 4x^{7}+3$, $y'=28x^{6}$. At the point $(4,0)$, $y'(4)=28\times4^{6}\neq13$.

Step4: Differentiate option C

For $y = 3x^{3}+4x$, $y'=9x^{2}+4$. At the point $x=-1$, $y'(-1)=9\times(-1)^{2}+4=9 + 4=13$. Also, when $x=-1$, $y=3\times(-1)^{3}+4\times(-1)=-3 - 4=-7\neq0$ (there is a mis - calculation in the original problem setup for the $y$ - value of the point, but the slope calculation is correct). The slope at $x = - 1$ is 13.

Step5: Differentiate option D

For $y = 5x^{7}+4$, $y'=35x^{6}$. At the point $(0,0)$, $y'(0)=35\times0^{6}=0\neq13$.