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# Core 3 maths watch

1. a) Find the equation of the normal to the curve y=e^-3x+2 at the point (0,3), and find the coordinates of the point where this normal crosses the x- axis.

b)Fine the region bounded by the curve with equation y=e^-3x+2, the coordinate axis and the line x=1.

2. (Original post by Carolyne2891)
a) Find the equation of the normal to the curve y=e^-3x+2 at the point (0,3), and find the coordinates of the point where this normal crosses the x- axis.

b)Fine the region bounded by the curve with equation y=e^-3x+2, the coordinate axis and the line x=1.

Then show us your process and we'll show you where you trip up. We're not doing the Q for you.
3. a) Find the derivative and plug in x=0 then find -1/that. Plug that into your point and find the equation. Set this to 0 and solve.
b) I don't believe integration is C3. However, this may be a thinking problem. If you know dy/dx = y then you know the integral of y must also be y. Use this concept and the definition of an integral (the opposite of a derivative) to solve the equation. Find the definite integral between the given limits.
4. dy/dx= -3e^-3x
which is over 1/(-3e^3x)
meaning m of normal is 3e^3x
y=e^-3x +2
0=e^-3x +2
not sure how to solve for x with logs,
but the answer is (-9,0) and equation 3y=x+9 ???
Not a clue
5. (Original post by Carolyne2891)
dy/dx= -3e^-3x
which is over 1/(-3e^3x)
meaning m of normal is 3e^3x
y=e^-3x +2
0=e^-3x +2
not sure how to solve for x with logs,
but the answer is (-9,0) and equation 3y=x+9 ???
Not a clue
It asks for the gradient at point (0,3) so you have to take x as 0 to find the gradient. Also, only e has a negative power so if you want to use that form it's -3/e^3x.
6. (Original post by Carolyne2891)
dy/dx= -3e^-3x
which is over 1/(-3e^3x)
meaning m of normal is 3e^3x
That's not right.

You now want the normal gradient the point where , thus construct the equation of your normal.
7. Thank you everyone!

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