f'(x)=4x^2 + 7x -9
f(x) = y = (4/3) x^3 + (7/2)x^2 - 9 x + constant
use point to get constant
6 = 0 + constant
The curve y = x/(1 + x^2) is called a serpentine. Find an equation of the tangent line to this curve at the point(2, 0.40).(Round the slope and y-intercept to two decimal places.)
The line that is normal to the curve x^2=2xy-3y^2=0 at(1,1) intersects the curve at what other point? Please help. Thanks in advance. We have x2=2xy - 3y2 = 0 Are there supposed to be 2 equal signs in this expression or is it x2 + 2xy - 3y2 = 0 ? I'll
Find an equation of the tangent line to the curve at the given point. y = 6 x sin x P= (pi/2 , 3pi) i know the slope of a tangent line is equal to the first derivative. For that I got 6xcosx + 6sinx but idk how to put that into the y-y1=m(x-x1) formula to
Use implicit differentiation to find an equation of the tangent line to the curve. sin(x+y)=4x−4y at the point (π,π). Tangent Line Equation:
The point P(7, −4) lies on the curve y = 4/(6 − x). (a) If Q is the point (x, 4/(6 − x)), use your calculator to find the slope mPQ of the secant line PQ (correct to six decimal places) for the following values of x. (i) 6.9 mPQ = (ii) 6.99 mPQ =
The curve y = x/(1 + x^2)is called a serpentine. Find an equation of the tangent line to this curve at the point (4, 0.24). Round the slope and y-intercept to two decimal places.)
Find the slope of the tangent line to the ellipse x^2/36 + y^2/49 =1 at the point (x,y). slope =_______ Are there any points where the slope is not defined? (Enter them as comma-separated ordered-pairs, e.g., (1,3), (-2,5). Enter none if there are no such
Find the slope m of the tangent to the curve y = 5/ sqrt (x) at the point where x = a > 0. what is the slope?
Show that the curve y=6x^3+5x-3 has no tangent line with slope 4. The answer key says that m=y'=18x^2+5, but x^2 is greater than or equal to 0 for all x, so m is greater than or equal to 5 for all x. I don't understand that x^2 is greater than or equal to
Find normals to the curve xy+2x-y=0 that are parallel to the line 2x+y=0 I have the answer: at(-1,-1), y=-2x-3, and at (3,-3), y=-2x+3 How do we get this? Thanks. xy + 2x - y=0 x dy/dx + y + 2 - dy/dx=0 dy/dx (x-1)= -y-2 dy/dx= - (y+2)/(x-1) The normal to
The curve with the equation y^2=5x^4-x^2 is called a kampyle of Eudoxus. Find an equation of the tangent line to this curve at the point (-1,2).
The point P(2,-1) lies on the curve y=1/(1-x) If Q is the point (x, 1/(1-x) find slope of secant line. these are the points 2, -1 1.5,2 1.9,1.111111 1.99,1.010101 1.999,001001 2.5,0.666667 2.1,0.909091 2.01,0.990099 2.001,0.999001 using the results from
1. use the definition mtan=(f(x)-f(x))/(x-a) to find the SLOPE of the line tangent to the graph of f at P. 2. Determine an equation of the tangent line at P. 3. Given 1 & 2, how would I plot the graph of f and the tangent line at P if : f(x)=x^2 +4,
-find the equation of the tangent line to the curve y=5xcosx at the point (pi,-5pi) -the equation of this tangent line can be written in the form y=mx+b where m= and b= -what is the answer to m and b?
graph the line with slope 1/2 passing through the point(-5,-2) find the slope of the line 5x+5y=3 write answer in simplest from consider the line 2x-4y=4 what is the slope of a line perpendicular to this line. what jis the slope of a line parallel to this
The slope of the tangent line to the curve y= 3x^3 at the point (-1,-3) is:? The equation of this tangent line can be written in the form y=mx+b where m is:? and where b is:?
For what values of a and b is the line 3x+y=b tangent to the curve y=ax^3 when x=–3?
1. Given the curve a. Find an expression for the slope of the curve at any point (x, y) on the curve. b. Write an equation for the line tangent to the curve at the point (2, 1) c. Find the coordinates of all other points on this curve with slope equal to
Consider the curve given by the equation y^3+3x^2y+13=0 a.find dy/dx b. Write an equation for the line tangent to the curve at the point (2,-1) c. Find the minimum y-coordinate of any point on the curve. the work for these would be appreciated i don't need
4. Given ln(x/y) + y^3 - 2x = -1. A. Find the equation of the normal line to the curve ln(x/y) + y^3 - 2x = -1 at the point (1,1). (I got -2x+3) B. Find the equation of a tangent line to the curve y=e^(x^2) that "also" passes through the point (1,0). You
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