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For the function
f(x)=-8x^(2)+9x+9, find the slope of the tangent line at
x=9.
Answer:

For the function $f(x)=-8 x^{2}+9 x+9$ , find the slope of the tangent line at $x=9$ . $\newline$ Answer:

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Find the slope of a tangent line using limits

Full solution

Q. For the function

f(x)=-8 x^{2}+9 x+9

, find the slope of the tangent line at

x=9

.

\newline

Answer:

Find Derivative: To find the slope of the tangent line to the function at a specific point, we need to find the derivative of the function, which gives us the slope of the tangent line at any point $x$ . The function is $f(x) = -8x^2 + 9x + 9$ . We will use the power rule for differentiation, which states that the derivative of $x^n$ is $n\cdot x^{(n-1)}$ .
Apply Power Rule: Differentiate the function with respect to $x$ . The derivative of $-8x^2$ is $-16x$ (using the power rule, the exponent $2$ becomes the coefficient and we subtract $1$ from the exponent). The derivative of $9x$ is $9$ (the derivative of $x$ is $1$ , so $9$ times $1$ is $9$ ). The derivative of the constant $9$ is $-8x^2$ $3$ (the derivative of a constant is always $-8x^2$ $3$ ). So, the derivative $-8x^2$ $5$ .
Evaluate Derivative: Now we need to evaluate the derivative at $x=9$ to find the slope of the tangent line at that point. $\newline$ Substitute $x=9$ into the derivative $f'(x) = -16x + 9$ . $\newline$ $f'(9) = -16(9) + 9$ .
Calculate Slope: Calculate the value of $f'(9)$ . $\newline$ $f'(9) = -16(9) + 9 = -144 + 9 = -135$ . $\newline$ So, the slope of the tangent line at $x=9$ is $-135$ .

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