Excel in math and science.

Join using Facebook Join using Google Join using email

Forgot password? New user? Sign up

Existing user? Log in

Your answer seems reasonable. Find out if you're right!

That seems reasonable. Find out if you're right!

Join using Facebook Join using Google Join using email

Forgot password? New user? Sign up

Existing user? Log in

Calculus

Curve Sketching

Concept Quizzes

Curve Sketching Warmup
Turning Points
Curve Sketching
Increasing / Decreasing Functions
Vertical Asymptotes
Relative Magnitude of Functions
Tangent to a Curve
Normal to a Curve

Challenge Quizzes

Curve Sketching: Level 3 Challenges
Curve Sketching: Level 4 Challenges

Normal to a Curve

Below is the graph of the curve $y=f(x).$ The green line is tangential to the curve at the point $P=(a,f(a)).$ If the red line is perpendicular to the green line and the equation of the red line is $y=-\frac{1}{5}x+15,$ what is the value of $f'(a)?$

Show explanation

View wiki

by Brilliant Staff

What is the normal to the curve $y=6\tan x + 4$ at the point $\left(\frac{\pi}{4}, 10\right)$ ?

\displaystyle{ y= -{12}x+\frac{\pi}{48}+{10} }

\displaystyle{ y= {6}x-\frac{\pi}{24} }

\displaystyle{ y= -\frac{1}{12}x+\frac{\pi}{48}+{10} }

\displaystyle{ y= \frac{1}{6}x-\frac{\pi}{24} }

Show explanation

View wiki

by Brilliant Staff

What is the equation of the line $l$ which passes through the point $(5,2)$ and is perpendicular to the line passing through the two points $(3,3)$ and $(9,4)?$

\displaystyle{y={-8}x-{37} }

\displaystyle{y= {7}x+{40} }

\displaystyle{y= {-6}x+{32} }

\displaystyle{y={9}x+{34} }

Show explanation

View wiki

by Brilliant Staff

What is the normal to the curve $y=2x^2 - 3x -15$ at the point $(3, -6)$ ?

\displaystyle{y= \frac{1}{9}x+\frac{17}{3} }

\displaystyle{y= {9}x+\frac{17}{3} }

\displaystyle{y= -{9}x-\frac{17}{3} }

\displaystyle{y= -\frac{1}{9}x-\frac{17}{3} }

Show explanation

View wiki

by Brilliant Staff

What is the normal to the curve $y=10{e}^x + {x}^2 +5$ at the point $(3,10{e}^3+14 )$ ?

\displaystyle{ y=-\frac{1}{{10}{e}^3+{6}}x+\frac{3}{{10}{e}^3+6}+{10}{e}^3+{14} }

\displaystyle{ y=\frac{1}{{10}{e}^3}x-\frac{3}{{4}{e}^3}-{8} }

\displaystyle{ y=-({10}{e}^2+{3})x+{10}{e}^3+{10}{e}^3+{14} }

\displaystyle{ y={10}{e}^3x+{4}{e}^3+{12}{e}^2+{7} }

Show explanation

View wiki

by Brilliant Staff