Answer to Question #291468 in Calculus for ask your question

Question

For each problem, find the equation of the line normal to the function at the given point. If the normal line is vertical line, indicate so. Otherwise, you answer should be in slope-intercept form.

1) y = -2x^2 - 4x at (-3, -6)

2) y = -x^3 + 2x^2 - 3 at (1, -2)

Answer 1

1) We take the function and derivate it: $y' = \frac{dy}{dx}= \frac{d}{dx}(-2x^2-4x)=-4x-4=-4(x+1)$

The slope m₁ of the tangent line will be: $y'(-3, -6)=-4(-3+1)=-4(-2)=8=m_1$

We need to use the relation to find the slope of the normal line m₂: $m_1m_2=-1 \to m_2 = -\frac{1}{m_1}=-\frac{1}{8}$

We use the point (-3, -6) to find the whole equation:

$y-y_0 = m(x-x_0) \to y-(-6)=-\frac{1}{8}(x-(-3)) \\ \to y+6=-\frac{x}{8}-\frac{3}{8} \to y = -\frac{x}{8}-\frac{3}{8}-6=-\frac{x}{8}-\frac{51}{8}$

2) We take the function and derivate it:

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

The slope m₁ of the tangent line will be: $y'(1, -2)=(1)(4-3\cdot 1)=1(1)=1=m_1$

We need to use the relation to find the slope of the normal line m₂: $m_1m_2=-1 \to m_2 = -\frac{1}{m_1}=-\frac{1}{1}=-1$

We use the point (1, -2) to find the whole equation:

$y-y_0 = m(x-x_0) \to y-(-2)=-1\cdot(x-1) \\ \to y+2=-x+1 \to y = -x-1$

In conclusion, the equations for the lines normal to the functions are:

1) $y = -\frac{x}{8}-\frac{51}{8}$

2) $y = -{x}-1$