Question #291468

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)


1
Expert's answer
2022-02-07T16:31:22-0500

1) We take the function and derivate it: y=dydx=ddx(2x24x)=4x4=4(x+1)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=m1y'(-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: m1m2=1m2=1m1=18m_1m_2=-1 \to m_2 = -\frac{1}{m_1}=-\frac{1}{8}


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

yy0=m(xx0)y(6)=18(x(3))y+6=x838y=x8386=x8518y-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=dydx=ddx(x3+2x23)y=3x2+4x=x(43x)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)(431)=1(1)=1=m1y'(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: m1m2=1m2=1m1=11=1m_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:

yy0=m(xx0)y(2)=1(x1)y+2=x+1y=x1y-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=x8518y = -\frac{x}{8}-\frac{51}{8}

2) y=x1y = -{x}-1


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS