If $f(x)=3 x^{2}+6 x+2$, find $f^{\prime}(-5)$, using the definition of derivative. $f^{\prime}(-5)$ is the limit as $x \rightarrow$ $\square$ of the expression $\square$
The value of this limit is $\square$
Use this to find the equation of the tangent line to the parabola $y=3 x^{2}+6 x+2$ at the point $(-5,47)$. The equation of this tangent line can be written in the form
$$
y=\square
$$
$\square$

Q: If $f(x)=3 x^{2}+6 x+2$, find $f^{\prime}(-5)$, using the definition of derivative. $f^{\prime}(-5)$ is the limit as $x \rightarrow$ $\square$ of the expression $\square$ The value of this limit is $\square$ Use this to find the equation of the tangent line to the parabola $y=3 x^{2}+6 x+2$ at the point $(-5,47)$. The equation of this tangent line can be written in the form \[ y=\square \] $\square$

∻Solution by Steps∻ ‖ step 1 ⋮ To find the derivative of $ f(x) = 3x^2 + 6x + 2 $ at $ x = -5 $, we use the definition of the derivative: $ f'(-5) = \lim_{x \to -5} \frac{f(x) - f(-5)}{x + 5} $ ‖ ‖ step 2 ⋮ We first calculate $ f(-5) $: $ f(-5) = 3(-5)^2 + 6(-5) + 2 = 75 - 30 + 2 = 47 $. Thus, the limit becomes $ \lim_{x \to -5} \frac{f(x) - 47}{x + 5} $ ‖ ‖ step 3 ⋮ Now we substitute $ f(x) $ into the limit: $ \lim_{x \to -5} \frac{(3x^2 + 6x + 2) - 47}{x + 5} = \lim_{x \to -5} \frac{3x^2 + 6x - 45}{x + 5} $ ‖ ‖ step 4 ⋮ We can factor the numerator: $ 3(x^2 + 2x - 15) = 3(x + 5)(x - 3) $. Thus, the limit simplifies to $ \lim_{x \to -5} \frac{3(x + 5)(x - 3)}{x + 5} = \lim_{x \to -5} 3(x - 3) $ ‖ ‖ step 5 ⋮ Evaluating the limit as $ x \to -5 $: $ 3(-5 - 3) = 3(-8) = -24 $. Therefore, $ f'(-5) = -24 $ ‖ ‖ step 6 ⋮ The slope of the tangent line at the point (-5, 47) is $ m = -24 $. We can use the point-slope form of the line: $ y - y_1 = m(x - x_1) $ where $ (x_1, y_1) = (-5, 47) $ ‖ ‖ step 7 ⋮ Substituting the values into the point-slope form: $ y - 47 = -24(x + 5) $. Simplifying gives $ y = -24x - 120 + 47 $ or $ y = -24x - 73 $ ‖ ∻Answer∻ ⚹ The derivative $ f'(-5) = -24 $ and the equation of the tangent line is $ y = -24x - 73 $ ⚹ ∻Key Concept∻ ⚹ The derivative represents the slope of the tangent line to the curve at a given point. ⚹ ∻Explanation∻ ⚹ We used the definition of the derivative to find the slope at $ x = -5 $ and then applied the point-slope form to derive the equation of the tangent line. ⚹◊What is the formula for the definition of the derivative using limits?⍭ Generate me a similar question◊

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the derivative of $ f(x) = 3x^2 + 6x + 2 $ at $ x = -5 $, we use the definition of the derivative: $ f'(-5) = \lim_{x \to -5} \frac{f(x) - f(-5)}{x + 5} $ ‖ ‖ step 2 ⋮ We first calculate $ f(-5) $: $ f(-5) = 3(-5)^2 + 6(-5) + 2 = 75 - 30 + 2 = 47 $. Thus, the limit becomes $ \lim_{x \to -5} \frac{f(x) - 47}{x + 5} $ ‖ ‖ step 3 ⋮ Now we substitute $ f(x) $ into the limit: $ \lim_{x \to -5} \frac{(3x^2 + 6x + 2) - 47}{x + 5} = \lim_{x \to -5} \frac{3x^2 + 6x - 45}{x + 5} $ ‖ ‖ step 4 ⋮ We can factor the numerator: $ 3(x^2 + 2x - 15) = 3(x + 5)(x - 3) $. Thus, the limit simplifies to $ \lim_{x \to -5} \frac{3(x + 5)(x - 3)}{x + 5} = \lim_{x \to -5} 3(x - 3) $ ‖ ‖ step 5 ⋮ Evaluating the limit as $ x \to -5 $: $ 3(-5 - 3) = 3(-8) = -24 $. Therefore, $ f'(-5) = -24 $ ‖ ‖ step 6 ⋮ The slope of the tangent line at the point (-5, 47) is $ m = -24 $. We can use the point-slope form of the line: $ y - y_1 = m(x - x_1) $ where $ (x_1, y_1) = (-5, 47) $ ‖ ‖ step 7 ⋮ Substituting the values into the point-slope form: $ y - 47 = -24(x + 5) $. Simplifying gives $ y = -24x - 120 + 47 $ or $ y = -24x - 73 $ ‖ ∻Answer∻ ⚹ The derivative $ f'(-5) = -24 $ and the equation of the tangent line is $ y = -24x - 73 $ ⚹ ∻Key Concept∻ ⚹ The derivative represents the slope of the tangent line to the curve at a given point. ⚹ ∻Explanation∻ ⚹ We used the definition of the derivative to find the slope at $ x = -5 $ and then applied the point-slope form to derive the equation of the tangent line. ⚹◊What is the formula for the definition of the derivative using limits?⍭ Generate me a similar question◊

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