Write the equations of all tangents to the curve y= -2x^2+12x-12 that pass thru (3,8)?

Question

Write the equations of all tangents to the curve y= -2x^2+12x-12 that pass thru (3,8)

Answer 1

Slope equation from derivative of y(x)
m = 12 - 4x

Slope equation through point
m = [ 8 - y(x) ] / [ 3 - x ]

Set them equal
[ 8 - y(x) ] / [ 3 - x ] = 12 - 4x
[ 8 - y(x) ] = [ 12 - 4x ] [ 3 - x ]
8 - y(x) = 36 - 24x + 4x^2
y(x) = 24x - 4x^2 - 28
-2x^2 +12x-12 = 24x - 4x^2 - 28
2x^2 + 16 = 12x
x^2 - 6x + 8 = 0
(x - 4) (x - 2) = 0

x = 4 and 2

y(x) at 4 and 2 is 4

m at 4 and 2 is -4 and 4, respectively

Can you make line equations from all this information?

Answer 2

y'=-4x+12. For a given value of x, y' can be found. From the equation above, the tangent line (TL) at any value of x has the SLOPE m = -4x+12. The TLs have the equation
y= (-4x+12) x + b
To pass through 3,8
b= 8 .
Then y= (-4X+12) x + 8
where X is the x-coordinate of the point on the curve, while x is a variable.

Answer 3

Differentiating, y' = -4x+12, which gives the slope of all such lines, so the general equation for the tangents will be y = (-4x+12)x + c, and at the point in question we will have 8 = (-4(3)+12)3 +c, which is trivially solved for c = 8 and gives us y = 8.

Answer 4

observe first that (3,8) is not on the parabola...

then let (a,b) be the point on the parabola whose tangent will pass through (3,8). [Note: (a,b) = (a, -2a²+12a-12)]

the slope of such a tangent is y'(a) = -4a + 12

then the equation of the line is...
(y-b)=m(x-a)

y + 2a² - 12a + 12 = (-4a + 12)(x-a) ... this is the unsimplified tangent line...

then one such (x,y) is the point (3,8).

8 + 2a² - 12a + 12 = (-4a + 12)(3-a)
2a² - 12a + 20 = 4a² - 24a + 36
2a² -12a + 16 = 0
a² - 6a + 8 = 0

thus a = 4 & a = 2

plug those values to the unsimplified tangent... and you have your tangent lines... §

at a = 2...
y - 4 = 4(x-2)
y = 4x - 4 ... the tangent line at the point (2,4)


at a = 4
y - 4 = (-4)(x-4)
y = -4x + 20 .. the tangent line at the point (4,4)

Answer 5

take the first deriative y'= -4x+12
the slope for any point along the curve is m= -4x+12
at 3 the slope is 0
now using point/intercept formula
y-y0=m(x-x0) here x0=3 yo=8 and m=0
y-8=0(x-3) y-8=0 y=8

tangent line y=8

Answer 6

-4x+ 20