calc help plzzzz?!?

Question

Find equations of both lines through the point (2,-3) that are tangent to the parabola y = x2 + x.
y =(smaller slope)
y = (larger slope)

I know you have to find the values for x and y where the slope to (2, -3) is the same as the slope of the tangent.

But how?

Answer 1

Let the coordinates where the tangent touches the parabola be (x,y)
Slope to (2,-3) = (y+3)/(x-2) = (x^2 + x + 3)/(x-2)

y = x^2 + x
dy/dx = 2x + 1

Both slopes must be equal
2x+1 = (x^2 + x + 3)/(x-2)
(2x+1)(x-2) = x^2 + x + 3
2x^2 - 3x - 2 = x^2 + x + 3
x^2 -4x - 5 = 0
(x+1)(x-5) = 0
Two values of x are -1 and 5

dy/dx = 2x + 1 ------- x=-1
= -1
y = -x+c
c = -1
x+y+1 = 0

dy/dx = 2x+1 ------- x=5
= 11
y = 11x+c
c = -25
11x-y-25=0

The 2 equations are
x+y+1=0
11x-y-25=0

Answer 2

Find equations of both lines through the point (2,-3) that are tangent to the parabola y = x² + x.

The slope of the lines at the point of tangency will be the same as the slope of the curve.

dy/dx = 2x + 1

Equation of the thru (2, -3).

y + 3 = m(x - 2)
x² + x + 3 = (2x + 1)(x - 2)
x² + x + 3 = 2x² - 3x - 2
0 = x² - 4x - 5
x² - 4x - 5 = 0
(x +1)(x - 5) = 0
x = -1, 5

y = x² + x = 0, 30

The first tangent line goes thru the points (2, -3) and (-1, 0).

m = ∆y/∆x = (0 + 3)/(-1 - 2) = 3/-3 = -1

y - 0 = -1(x + 1)
y = -x - 1

The second tangent line goes thru the points (2, -3) and
(5, 30).

m = ∆y/∆x = (30 + 3)/(5 - 2) = 33/3 = 11

y - 30 = 11(x - 5)
y - 30 = 11x - 55
y = 11x - 25

The two tangent lines are:

y = -x - 1
y = 11x - 25

Answer 3

The points on the parabola are (x, x^2 + x)

so find the slope of the line from (2, -3) to (x, x^2 + x) to get:

slope = (x^2 + x + 3)/(x-2) (i just used slope formula)

But the slope of the tangent line to the parabola must equal 2x + 1 (this is the derivative)

So set those equal to each other and solve. You get x =5 or x = -1. These are the x values of the points on the parabola. Use them to find the y values. then just find the equation of the lines from (2, -3) to each of those two points we just found. That's it.

Answer 4

Let (x, x^2+x) be any point on the parabola.
m = (x^2+x+3)/(x-2) = y' = 2x+1
Therefore, we have
x^2+x+3 = 2x^2-3x-2
x^2-4x-5 = 0
(x-5)(x+1) = 0
x = 5, y = 5^2+5 = 30, y' = 2(5)+1 = 11
y -30 = 11(x-5)
or
x = -1, y = (-1)^2 - 1 = 0, y' = 2(-1)+1 = -1
y = -(x+1)

Answer 5

i'll assume yo mean y= x^2 +x

first differentiate the function

y' = 2x +1

then find the gradiant of the function by sub in the point

2(2)+1 = 5

thus the eqn of one line will be -3 = 5(2) +c u can work out eqn from there
the other eqn will be perpendicular to that line