Question

find the slope of the tangent line to the graph of the given function at the given value of x. find the equation of the tangent line. y = x^4 - 5x^3 + 9; x = 2
how would the slope of a tangent line be determined with the given information?
a. substitute values of x into the equation and solve for y. plot the resulting points to find the linear equation.
b. substitute values of y into the equation and solve for x. plot the resulting points to find the linear equation.
c. set the derivative equal to zero and solve for x.
d. substitute 2 for x into the derivative of the function and evaluate.

Explanation:

Step1: Find the derivative of the function

The power - rule for differentiation is $\frac{d}{dx}(x^n)=nx^{n - 1}$. Given $y=x^{4}-5x^{3}+9$, then $y'=\frac{d}{dx}(x^{4}-5x^{3}+9)=4x^{3}-15x^{2}$.

Step2: Find the slope of the tangent line

To find the slope of the tangent line at $x = 2$, substitute $x = 2$ into the derivative. So, $m=y'(2)=4(2)^{3}-15(2)^{2}=4\times8 - 15\times4=32-60=-28$.

Step3: Find the y - coordinate of the point on the curve

Substitute $x = 2$ into the original function $y=x^{4}-5x^{3}+9$. Then $y=(2)^{4}-5(2)^{3}+9=16-40 + 9=-15$.

Step4: Find the equation of the tangent line

Use the point - slope form of a line $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(2,-15)$ and $m=-28$. So, $y+15=-28(x - 2)$, which simplifies to $y+15=-28x + 56$, or $y=-28x+41$.

For the multiple - choice question:
The slope of the tangent line to a function $y = f(x)$ at a given value of $x=a$ is found by substituting $a$ into the derivative of the function. So the correct way to determine the slope of the tangent line with the given information is to substitute 2 for $x$ into the derivative of the function and evaluate. The answer is D.

Answer:

The slope of the tangent line is $-28$.
The equation of the tangent line is $y=-28x + 41$.
D. Substitute 2 for x into the derivative of the function and evaluate.