Question

Find the gradients of the lines that pass through the point (1, 7) and are tangent to the parabola y = (2 − x)(1 + 3x)

Answer 1

@oldrin.bataku help please

Answer 2

expanded -3x^2 + 5x + 2

Answer 3

define the derivative now

Answer 4

-6x + 5

Answer 5

then that will define the slope of the tangent line for some x=a;

attach that to the given point

Answer 6

m = -1?

Answer 7

no, m=(-6a+5) attached to the point 1,7

Answer 8

? what do you mean attach it?

Answer 9

do you recall how to form a line given a slope and a point?

Answer 10

y-y1 = m(x-x1)

Answer 11

correct, use that ....

Answer 12

y-7 = (5-6a)(x-1)

Answer 13

now, for some point (a,b); the parabola and the line equal ... so equate them.

Answer 14

x^2 - 10x + 6ax - 6a - 5 = 0

Answer 15

this is where my thought goes; for some x=a

-3a^2 + 5a + 2 = (-6a+5)a + 6a +2

Answer 16

solve for a

Answer 17

using B^2 - 4ac right?

Answer 18

if you want to use the quadratic formula, then thats fine.

Answer 19

i have gotten 3a^2 - 8a + 10
I think i went wrong somewhere

Answer 20

-3a^2 + 5a + 2 = (-6a+5)a + 6a +2

-3a^2 + 5a = (-6a+5)a + 6a

-3a^2 = -6a^2+5a + a

3a^2 -6a = 0

3a(a -2) = 0, when a=0, or a=2

Answer 21

ok now we sub it into y=mx+b?
or something like that?

Answer 22

well, we already have the line equation; we just need to fill in our a values

y = (5-6a)(x-1)+7

y1 = 5(x-1)+7
y2 = -7(x-1)+7

http://www.wolframalpha.com/input/?i=y%3D-3x%5E2+%2B+5x+%2B+2%2C+y-7+%3D+%285%29%28x-1%29%2C+y-7+%3D+%285-6%282%29%29%28x-1%29

Answer 23

So gradients would be 5 and -7?

Answer 24

yep

Answer 25

thanks for your help

Answer 26

youre welcome