Write the equation of the tangent line to the curve at the indicated point. $$ y=2 x^{2}-3 x+6 \text { at } x=-2 $$

Question

UpStudy AI · Accepted Answer

To find the equation of the tangent line to the curve $$ y = 2x^{2} - 3x + 6 $$ at the point where $$ x = -2 $$, follow these steps:

### 1. Find the Point of Tangency
First, determine the $$ y $$-coordinate when $$ x = -2 $$:

$$
y = 2(-2)^{2} - 3(-2) + 6 = 2(4) + 6 + 6 = 8 + 6 + 6 = 20
$$

So, the point of tangency is $$(-2, 20)$$.

### 2. Determine the Slope of the Tangent Line
Next, find the derivative of $$ y $$ with respect to $$ x $$ to get the slope of the tangent line:

$$
\frac{dy}{dx} = \frac{d}{dx}(2x^{2} - 3x + 6) = 4x - 3
$$

Evaluate the derivative at $$ x = -2 $$:

$$
\left. \frac{dy}{dx} \right|_{x=-2} = 4(-2) - 3 = -8 - 3 = -11
$$

So, the slope $$ m $$ of the tangent line is $$-11$$.

### 3. Write the Equation of the Tangent Line
Using the point-slope form of a line:

$$
y - y_{1} = m(x - x_{1})
$$

Plug in the point $$(-2, 20)$$ and the slope $$-11$$:

$$
y - 20 = -11(x - (-2)) \
y - 20 = -11(x + 2)
$$

Simplify to slope-intercept form:

$$
y = -11x - 22 + 20 \
y = -11x - 2
$$

### Final Answer

The equation of the tangent line is

$$
y = -11\,x\ -\ 2
$$
The equation of the tangent line is $$ y = -11x - 2 $$.

Write the equation of the tangent line to the curve at the indicated point. \[ y=2 x^{2}-3 x+6 \text { at } x=-2 \]