# Calculus derivative

Find the slope and an equation of the tangent line to the graph of the function f at the specified point.

f(x)=-1/3x^2+5x+5: (-1, -1/3)

y=-2/3x -1

I re-worked the problem and got f'(x)=5, y=5x+ 14/3

This is from a multiple choice test and this is not an answer.

Can someone check my work?

1. 👍 0
2. 👎 0
3. 👁 169
Now use the value of x of the given point to find the slope
slope = (-2/3)(-1) + 5
= 2/3 + 5 = 17/3

equation is
y = (17/3)x + b
sub in the point
-1/3 = (17/3)(-1) + b
16/3 = b

equation: y = (17/3)x + 16/3

1. 👍 0
2. 👎 0
2. Thanks. I was missing the step where you plug in the -1 to get slope.

1. 👍 0
2. 👎 0

## Similar Questions

1. ### Math(check)

1)Find the range of the relation [(-1,4)},(2,5),(3,5)}.Then determine whether the relation is a funtion. 4,5,5 is the range 4,5 is a function 2)Find f(-1),if f(x)= x^2-6x/x+2 (-1)^2-6(-1)/-1+2 1+6/-1+2 7/1 =7 3)Find f(a),if f(t)=

asked by erin on July 26, 2007
2. ### calculus

1. Which of the following expressions is the definition of the derivative of f(x) = cos(x) at the point (2, cos(2))? 2. Find the derivative of f(x) = |x + 2| at the point (1, 3) 3. Find f '(x) for f(x) = -2x3 + 3x2 - x + 15. 4.

asked by mock on January 15, 2015
3. ### I would like to understand my calc homework:/

Consider the differential equation given by dy/dx=(xy)/(2) A) sketch a slope field (I already did this) B) let f be the function that satisfies the given fifferential equation for the tangent line to the curve y=f(x) through the

asked by Amber on March 27, 2013
4. ### Math

Find the equation of the line tangent to curve x=sec(t), y=tan(t), at t=pi/6. I found th slope(derivative of the two which was csc(t). Plugged in Pi/6 and got 2 as the slope. Now how do I find the equation of the tangent? There

asked by Beth on December 17, 2015
1. ### Calculus - Functions?

#1. A cubic polynomial function f is defined by f(x) = 4x^3 +ax^2 + bx + k where a, b and k are constants. The function f has a local minimum at x = -1, and the graph of f has a point of inflection at x= -2 a.) Find the values of

asked by Amy on February 21, 2011
2. ### Math (Secant Lines)

Consider the function f(x)=sqrt(x) and the point P(4,2) on the graph of f? -Consider the graph f with secant lines passing through p(4,2) and Q(x,f(x)) for x-values 1, 3, and 5. -Find the slope of each secant line -Use the results

asked by Ray on October 1, 2016
3. ### Calculus

1. a) For the Function and point below , Find f’(a). b) Determine the equation of the line tangent to the graph of f at (a,f(a)) for the given value of f(x) = 4x2+2x, a =1 F’(a) = y = 2. For the function find f’ using the

asked by Tired on February 11, 2012
4. ### calculus

Let g be a function that is defined for all x, x ≠ 2, such that g(3) = 4 and the derivative of g is g′(x)=(x^2–16)/(x−2), with x ≠ 2. Find all values of x where the graph of g has a critical value. For each critical

asked by Anonymous on December 1, 2016
1. ### Calculus

Find the slope m of the tangent line to the graph of the function at the given point and determine an equation of the tangent line. f(x) = 7 x - 2 x^2 text( at ) \(-1,-9\) m = y =

asked by George A.J on January 31, 2011
2. ### math

Find the slope m of the tangent line to the graph of the function at the given point and determine an equation of the tangent line. f(x)=7x-5x^2 at (-2,-34) m = ?? y = ?? .

asked by Kimberly on March 4, 2013
3. ### Calculus

The equation dy/dx = -6x^2/y gives the slope at any point on the graph of f(x). The range of f(x) is [0, infinity] and f(1) = 2. A. Find the equation of the tangent line to f(x) at the point (1,2). B. Write the function f(x). C.

asked by Anonymous on April 23, 2016
4. ### Math (Calculus) (mean value theorem emergency)

Consider the graph of the function f(x)=x^2-x-12 a) Find the equation of the secant line joining the points (-2,-6) and (4,0). I got the equation of the secant line to be y=x-4 b) Use the Mean Value Theorem to determine a point c

asked by Ray on November 19, 2016