Graham Bowman
08/03/2024 · High School
Find an equation of the line tangent to the graph of \( f(x)=-2-8 x^{2} \) at \( (-2,-34) \)
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Tutor-Verified Answer
Step-by-step Solution
To find the equation of the tangent line to the graph of the function \( f(x) = -2 - 8x^2 \) at the point \( (-2, -34) \), we need to follow these steps:
1. **Find the derivative of the function** to determine the slope of the tangent line at the given point.
2. **Evaluate the derivative at \( x = -2 \)** to find the slope at that point.
3. **Use the point-slope form of the equation of a line** to write the equation of the tangent line.
### Step 1: Find the derivative of \( f(x) \)
The function is given by:
\[
f(x) = -2 - 8x^2
\]
To find the derivative \( f'(x) \), we differentiate \( f(x) \):
\[
f'(x) = \frac{d}{dx}(-2) + \frac{d}{dx}(-8x^2) = 0 - 16x = -16x
\]
### Step 2: Evaluate the derivative at \( x = -2 \)
Now we evaluate the derivative at \( x = -2 \):
\[
f'(-2) = -16(-2) = 32
\]
So, the slope of the tangent line at the point \( (-2, -34) \) is \( 32 \).
### Step 3: Use the point-slope form of the equation of a line
The point-slope form of the equation of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line.
Substituting \( m = 32 \), \( x_1 = -2 \), and \( y_1 = -34 \):
\[
y - (-34) = 32(x - (-2))
\]
This simplifies to:
\[
y + 34 = 32(x + 2)
\]
Now, distribute the \( 32 \):
\[
y + 34 = 32x + 64
\]
Finally, isolate \( y \):
\[
y = 32x + 64 - 34
\]
\[
y = 32x + 30
\]
### Conclusion
The equation of the line tangent to the graph of \( f(x) = -2 - 8x^2 \) at the point \( (-2, -34) \) is:
\[
\boxed{y = 32x + 30}
\]
Quick Answer
The equation of the tangent line is \( y = 32x + 30 \).
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