Find the question of the tangent line to the function f(x)=x^2+x-2 at the point (-1,-2)?

find the equation of the tangent line to the fuction f(x)= x^2 + x - 2 at the point (-1,-2).





soo i found the derivative.. then do I just plug in the points?


so 2x+1 .. and plug in -1?





the sketch the curves y=x^2 +1 and y=-x^2 on the same axes. Find the equations of the lines that are tangent to both curves simultaneously.





Then.. the straight line tangent to y=x^3 at x=a crosses the y=axis at the point (0,2). Find the value of a.





Can someone help me? im a bit confused.. %26gt;_%26lt;

Find the question of the tangent line to the function f(x)=x^2+x-2 at the point (-1,-2)?
1. If y = f(x) = x^2 + x - 2, at (x, y) = (a, a^2 + a - 2) [which IS on the curve], the tangent is





y - (a^2 + a - 2) = (2a + 1)(x - a), that is





y = (2a + 1)x + (a^2 + a - 2) - a(2a + 1), or





y = (2a + 1)x - a^2 - 2. ......(M)





[CHECK: at x = a, y = a^2 + a - 2. which is correct.]





Since this has to pass through (-1, -2),





- 2 = - (2a + 1) - a^2 - 2, or a^2 + 2a + 1 = 0, that is (a + 1)^2 = 0,





so that a = - 1. ......(N)





So the desired tangent line is found by inserting a = - 1 from (N) into eqn (M), that is:





y = - x - 3.





QED





2. The tangent to y = x^2 + 1 at x = a has a slope 2a. It is therefore





y - (a^2 + 1) = 2a(x - a), or y = 2ax + a^2 + 1 - 2a^2 or





y = 2ax + 1 - a^2. ......(A)





The tangent to the curve y = - x^2 at x = b is y +b^2 = -2b(x - b), or y = -2bx + 2b^2 - b^2 , that is





y = - 2bx + b^2. ......(B)





So if these two lines are the same, the SLOPES must be equal, that is





2a = - 2b, that is a = - b, ......(C)





and the CONSTANT terms must be equal, that is





1 - a^2 = b^2. ......(D)





So, since a = - b, a^2 = b^2 and therefore by (D), 1 = 2a^2 (or 2b^2), so that





a = +/- sqrt (1/2), and correspondingly, b = -/+ sqrt(1/2).





So the TWO equations to the common tangents are:





y = +/- sqrt(2)x + 1/2.





QED





[These two common tangents both pass through (x, y) = (0, 1/2, as is OBVIOUS by symmetry, and have slopes of +/- sqrt(2).]





3. The straight line tangent to y = x^3 at x = a has a slope 3a^2. Its equation is therefore:





y - a^3 = 3a^2(x - a), or





y = 3a^2 x - 2a^3.





If it crosses the y-axis at (0, 2), 2 = 0 - 2 a^3, therefore





a = - 1.





QED





Live long and prosper.
Reply:1. find the question of the tangent line to the function f(x)=x^2+x-2 at the point (-1,-2)?


f'(x) = 2x+1


f(-1) = -2+1 = -1


y+2 = -(x+1),


or


y = -x-3, the equation of the tangent line





2. sketch the curves y=x^2 +1 and y=-x^2 on the same axes. Find the equations of the lines that are tangent to both curves simultaneously.


Let the tangent point on y = x^2+1 be (x1,x1^2+1), and the tangent point on y = -x^2 be (x2, -x2^2).


y' = 2x1 = -2x2 =%26gt; x1 = -x2


slope = (x1^2+1+x2^2)/(x1-x2) = (2x1^2+1)/(2x1) = 2x1


2x1^2+1 = 4x1^2


2x1^2 = 1


x1 = ±1/√2


y1 = 1/2+1 = 3/2


y' = 2x1 = ±√2


The equations of the lines are


y - 3/2 = ±√2(x -/+ 1/√2), point-slope form


or


y = ±√2x+(1/2), slope-intercept form





3. the straight line tangent to y=x^3 at x=a crosses the y=axis at the point (0,2). Find the value of a.


The tangent point is at (a, a^3)


slope = (2-a^3)/(0-a) = 3a^2, since y' = 3x^2.


2-a^3 = -3a^3


a^3 = -1


a = -1


The equation of the tangent line is: y = 3x+2.


------------


Dr Spock,


You changed your answer repeatedly, didn't you? You made the same error by misreading a negative sign as I did yesterday.


For the last problem, you did wrong again. "a=1" is not a right answer.





OK, you changed your answer again to "a = -1" after I posted "a=1" is not a right answer.