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Find the slope of the tangent line to the graph of $x = -4y^2 + 8y + 3$ at $(3,2)$ .

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Find tangent lines using implicit differentiation

Full solution

Q. Find the slope of the tangent line to the graph of

x = -4y^2 + 8y + 3

at

(3,2)

.

Find Derivative of $x$ : First, we need to find the derivative of $x$ with respect to $y$ to get the slope of the tangent line. Since $x = -4y^2 + 8y + 3$ , differentiate each term with respect to $y$ .
$\frac{dx}{dy} = \frac{d}{dy}(-4y^2) + \frac{d}{dy}(8y) + \frac{d}{dy}(3)$
$\frac{dx}{dy} = -8y + 8 + 0$
Substitute $y = 2$ : Next, substitute $y = 2$ into the derivative to find the slope at the point $(3,2)$ . $\newline$ Slope = $-8(2) + 8$ $\newline$ Slope = $-16 + 8$ $\newline$ Slope = $-8$

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