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Chapter 3: Problem 55

Tangent line to a parabola \(\quad\) Does the parabola \(y=2 x^{2}-13 x+5\) have a tangent line whose slope is -1? If so, find an equation for the line and the point of tangency. If not, why not?

Short Answer

Expert verified
Yes, the tangent line with slope -1 exists: it is \(y = -x - 13\).

Step by step solution

01

Find the Derivative

To determine if the parabola has a tangent line with a slope of \(-1\), we first need to find the derivative of the parabola. The derivative of the function \(y = 2x^2 - 13x + 5\) gives the slope of the tangent line at any point \(x\). The derivative is computed as follows: \[\frac{dy}{dx} = \frac{d}{dx}(2x^2) - \frac{d}{dx}(13x) + \frac{d}{dx}(5) = 4x - 13.\]
02

Set Derivative Equal to -1

Now, we set the derivative equal to -1 to find the point where the slope of the tangent line is \(-1\). \[4x - 13 = -1.\] solve for \(x\):\[4x = 12,\]\[x = 3.\]
03

Find the Point of Tangency

Substitute the value of \(x\) back into the original equation to find the corresponding \(y\)-value of the point of tangency:\[y = 2(3)^2 - 13(3) + 5 = 18 - 39 + 5 = -16.\]Therefore, the point of tangency is \((3, -16)\).
04

Equation of the Tangent Line

Now that we have a point on the tangent line \((3, -16)\) and the slope \(-1\), we use the point-slope form of a line to write the equation:\[y - y_1 = m(x - x_1),\]\[y + 16 = -1(x - 3),\]\[y + 16 = -x + 3,\]\[y = -x - 13.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative Finding
To determine the slope of the tangent line to a parabola at any given point, the first step is to compute the derivative of the parabola's equation. Let's focus on the given parabola, described by the equation:
\[ y = 2x^2 - 13x + 5. \]
The derivative, denoted by \( \frac{dy}{dx} \), is found by differentiating each term of the polynomial:
  • The derivative of \( 2x^2 \) is \( 4x \) because the power rule says to multiply the current exponent (2) by the coefficient (2), and decrease the exponent by 1.
  • The derivative of \(-13x\) is \(-13\) since the derivative of \(x\) with respect to \(x\) is 1.
  • The constant 5 becomes 0 because constants have no rate of change.
As a result, the derivative is:\[ \frac{dy}{dx} = 4x - 13. \]This formula gives the slope of the tangent line at any point \(x\) on the curve.
Slope Calculation
Calculating the slope of the tangent line involves using the derivative we have found. For our parabola, the derivative
\[ 4x - 13 \]
represents the slope at any point \(x\) on the curve. To find where this slope equals a specific value, say \(-1\), we simply set the derivative equal to \(-1\):
\[ 4x - 13 = -1. \]
Solving this equation, we first add 13 to both sides to isolate terms with \(x\):\[ 4x = 12. \]Then, divide both sides by 4 to solve for \(x\):
\[ x = 3. \]
This tells us that at \(x = 3\), the slope of the tangent line is \(-1\), exactly matching the slope we were looking for.
Point of Tangency
Identifying the exact location where the tangent line meets the parabola, known as the point of tangency, requires substituting the \(x\) value found from the slope calculation back into the original parabola equation. We found that at \(x = 3\), the slope is \(-1\). Now, let's find the corresponding \(y\) value:
\[ y = 2(3)^2 - 13(3) + 5. \]
Calculating step-by-step:
  • \(2(3)^2 = 2 \cdot 9 = 18\).
  • \(-13 \cdot 3 = -39\).
  • Add constants: \(18 - 39 + 5 = -16.\)
Therefore, the point of tangency is \((3, -16)\). This is where the line touches the parabola without crossing it.
Equation of Tangent Line
Once the point of tangency is known, \((3, -16)\), and the slope \(-1\) is confirmed, writing the equation of the tangent line is straightforward using the point-slope form:
\[ y - y_1 = m(x - x_1), \]where \((x_1, y_1)\) is the point of tangency and \(m\) is the slope. Plug in the values:
  • \(y + 16 = -1(x - 3)\)
  • Distribute the slope: \(y + 16 = -x + 3\).
  • Isolating \(y\), we get \(y = -x - 13\).
This equation, \(y = -x - 13\), represents the tangent line to the parabola at the point \( (3, -16) \). It describes a line that gently just grazes the parabola, never cutting across.

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