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 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.
the slope of the tangent line is - 28.
the equation of the line is . (type an equation. type your answer in slope - intercept form.)

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 $y^\prime=\frac{d}{dx}(x^{4}-5x^{3}+9)=4x^{3}-15x^{2}$.

Step2: Evaluate the derivative at $x = 2$

Substitute $x = 2$ into $y^\prime$. $y^\prime(2)=4\times2^{3}-15\times2^{2}=4\times8 - 15\times4=32-60=-28$. This is the slope of the tangent line.

Step3: Find the $y$-coordinate when $x = 2$

Substitute $x = 2$ into the original function $y=x^{4}-5x^{3}+9$. $y(2)=2^{4}-5\times2^{3}+9=16 - 40 + 9=-15$.

Step4: Use the point - slope form to find the equation of the tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(2,-15)$ and $m=-28$. So $y+15=-28(x - 2)$.

Step5: Rewrite in slope - intercept form

Expand the right - hand side: $y+15=-28x + 56$. Then $y=-28x+41$.

Answer:

$y=-28x + 41$