HELP!!! maths assessment (2 Viewers)

[Cattle]

I have a maths assessment due soon and i have no idea how to do half of the questions can someone please help me?

1. Find the equation of the tangents to the parabola y= x^2 -5x+6 at the points where it cuts the x- axis. where do these tangents intersect?

2. Given that f(x)= x^2+2x, find the exact values of m for which f'(m)= f(m)

Thanks

x^2 means x squared didnt know how else to write it sorry

 

[pLuvia]

1.

y= x²-5x+6
y'= 2x-5

Where it cuts x axis
x intercepts = 3 and 2
At x=3
y' = 1
At x=2
y'= -1

Eqn of tangents

y = x-3
&
y = -x+2

Solve simultaneously
x-3=-x+2
x = 5/2
y = -1/2

Pt of intersection (5/2,-1/2)

2.
f(m) = m²+2m
f'(m) = 2m+2

If f'(m)= f(m)
m²+2m = 2m+2
m² = 2
m = +sqrt 2

Edit: Mistake

 

[Holsworthy]

(1) the parabola cuts x-axis at x=2, 3 ((x-2)(x-3) = 0). The gradient of tangents is m
= 2x-5, so m1=-1, m2 = 1. So equation1: y = -x + 2 and equation2 y = x - 3. They meet at (5/2, -1/2) since -x + 2 = x - 3 => x = 5/2, and y therefore is -1/2 from any of
these two equations.

(2) f'(x) = 2x+ 2, so f'(m) = 2m + 2 = f(m)=m^2+2m => x = + & - sqrt(2)

Hope this helps

 

[Cattle]

Thankyou so much

 

[Cattle]

i dont want to be a pain but do you know how to do this question?

1. Find the following indefinite intergrals: (all meant to have integrating sign in front of them)
a) (12x^2 - 4) dx

b)(x^2 - squareroot x)^2 dx

c) 2du

d) x^2- 4 (top line over the 2nd)
x- 2

e) (3- 2x) ^5 dx

f) (x+ 1)(x-1)dx
x x (1's r meant to be over the x's)


Thanks

 

[Holsworthy]

I'd love to help you, but I need to go to a Xmas party now. Can I attend those questions for you later?

 

[Cattle]

yeh if you can answer them later it would be greatly appreciated

 

[webby234]

1a) 4x^3 - 4x + c

b)Int[ x^4 + x - 2x^(2.5)]
= 1/5x^5 +1/2x^2 - 4/7x^(3.5) + c

c)2u +c

 

[webby234]

d) = Int(x+2)(x-2)/(x-2)
=Int(x+2) x does not equal 2

= 1/2x^2 + 2x +c (x does not equal 2)

 

[webby234]

e) (3-2x)^6/(-12)

= -(3-2x)^6 /12

f) (x+1/x) (x-1/x)

= x^2 + x/x-x/x -1/x^2

=x^2 - x^(-2)

=(1/3)x^3 + x^-1

= (x^3)/3 + 1/(3x)

 

[Cattle]

thanks that helps a lot