question 1 (1 point)\nuse implicit differentiation to find the slope of the line tangent to the curve 4x² +…

Explanation:

Step1: Differentiate each term

Differentiate $4x^{2}+2x + xy=2$ with respect to $x$. The derivative of $4x^{2}$ is $8x$ using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$. The derivative of $2x$ is $2$. For the term $xy$, use the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u = x$ and $v = y$. So $\frac{d}{dx}(xy)=y+x\frac{dy}{dx}$. The derivative of the constant $2$ is $0$. We get $8x + 2+y+x\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Isolate $\frac{dy}{dx}$: [ \begin{align*} x\frac{dy}{dx}&=-8x - 2 - y\ \frac{dy}{dx}&=\frac{-8x - 2 - y}{x} \end{align*} ]

Step3: Substitute the point $(2,-9)$

Substitute $x = 2$ and $y=-9$ into $\frac{dy}{dx}$: [ \begin{align*} \frac{dy}{dx}&=\frac{-8\times2-2-(-9)}{2}\ &=\frac{-16 - 2 + 9}{2}\ &=\frac{-18 + 9}{2}\ &=\frac{-9}{2} \end{align*} ]

Answer:

$\frac{-9}{2}$