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Chapter 6: Problem 59

Find the equation of the tangent line to the parabola at the given point. $$y=-2 x^{2},(-1,-2)$$

Short Answer

Expert verified
The equation of the tangent line to the parabola \(y = -2x^{2}\) at the point \(-1,-2\) is \(y = 4x + 6\).

Step by step solution

01

Find The Derivative

First, take the derivative of \(y = -2x^{2}\). The derivative, \(y'\), is obtained using the power rule, which is \(nx^{n-1}\). So, the derivative is \(y'=-4x\).
02

Calculate The Derivative At The Point

Now, evaluate \(y'\) at the point \(-1, -2\). This will give the slope of the tangent line at that point. By substituting \(-1\) for \(x\) in \(y'\), you get \(y'(-1)=-4*(-1)=4\). The slope of the tangent line is thus \(4\).
03

Use Point-Slope Form to Get the Equation

Next, use the point-slope form of the line equation, which is \(y - y1 = m(x - x1)\), where \(m\) is the slope and \((x1, y1)\) is the point on the line, to find the equation of the tangent line. Substitute \(4\) for \(m\), \(-1\) for \(x1\), and \(-2\) for \(y1\). Then the equation is: \(y - (-2) = 4(x - (-1))\). Simplifying this, we get \(y = 4x + 6\). This is the equation of the tangent line to the parabola at the point \(-1,-2\).

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