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Find the slope of the tangent line to the graph of $-y^2 + 5y + 17 = x$ 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

-y^2 + 5y + 17 = x

at

(3,-2)

.

Find Derivative: First, we need to find the derivative of the given equation to find the slope of the tangent line. The equation is $-y^2 + 5y + 17 = x$ . We'll differentiate implicitly with respect to $x$ . $\frac{d}{dx}(-y^2 + 5y + 17) = \frac{d}{dx}(x)$ $-2y\frac{dy}{dx} + 5\frac{dy}{dx} = 1$ $\frac{dy}{dx}(-2y + 5) = 1$ $\frac{dy}{dx} = \frac{1}{(-2y + 5)}$
Implicit Differentiation: Next, substitute $y = -2$ into the derivative to find the slope at the point $(3, -2)$ . $\frac{dy}{dx} = \frac{1}{(-2(-2) + 5)}$ $\frac{dy}{dx} = \frac{1}{(4 + 5)}$ $\frac{dy}{dx} = \frac{1}{9}$

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