Question

I am going to post a picture of the question I am stuck on now, I am stuck on 6 (ii). Please help!

Answer 1

solve for x^2+4=4x and you have X^2-4x+4=0 implies that (x-2)^2=0 implies that x=2
now put x=2 in y=4x and you have y=8
so (2,8) is the point of intersection of the curves y=x^2+4 and y=4x

Answer 2

oh sorry I did the wrong sum
OK lets do 6 (ii) now.
dy/dx=8x^2-30x+42
at (2,40) dy/dx=14
so the required tangent must have slope 14
we solve for 8x^2-30x+42=14 implies 8x^2-30x+28=0 this gives x=2,7/4 but x=2 corresponds to the given point. so we take x=7/4 and putting x=7/4 in the given curve we have y=1225/32
so the required coordinates is (7/4,1225/32)

Answer 3

hmm I am afraid the answers at the back of the book say its (3,45)

Answer 4

But I see a mistake I madefrom your maths there. I thought slope was 6 for some reason

Answer 5

Oh wait, is it 6?

Answer 6

slope is 14 for sure...

Answer 7

6 (ii). IF (a,b) is another point on the curve for which the tangent is parallel to the one at (2,40), then \[\frac{dy}{dx}|_{x=a}=\frac{dy}{dx}|_{x=2}\]dy/dx

Answer 8

then solve for the value of \(a\). you should get two values since the equation is quadratic.

Answer 9

Im sorry, I am not quite following how do I solve for a?

Answer 10

x coordinate is 7/4 for sure... It cannot be 3.. you check once again

Answer 11

\[\frac{dy}{dx}=6x^2-30x+42\]
the slope at \((a,b)\) is \(6a^2-30a+42\)
the slope at \((2,40)\) is \(6\times 2^2-30\times 2+42\)
because the two tangents are parallel,
\[6a^2-30a+42=6\times 2^2-30\times 2+42\]
the equation is quadratic

Answer 12

Ok, I will try that quickly

Answer 13

now I see where I did wrong.. its 6x^2-30x+42

Answer 14

I am getting the right thing now! Thank you for all the help!

Answer 15

You're welcome

Answer 16

And thanks Jango_IN_DTOWN too. :D

Answer 17

ok so 6x^2-30x+42 at x=2 is 6
so we have to solve for 6x^2-30x+42=6
implies 6x^2-30x+36=0 implies x^2-5x+6=0 implies x=2,3 but x=2corresponds to the given value so we take x=3 at x=3, from the given curve we have y=45 so th required point of intersection is (3,45)