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# P2 Question Help watch

1. The function f is definted for all real values of x by

f(x) = |2x - 3| -1

Can somebody explain to me why the graph crosses the x axis at (1,0) and (2,0)

And the Y axis at (0,2)

I know the vertex is at -1. This has never been taught to us and I can't seem to figure them out on the exam papers. Any help would be much appreciated.

Thanks

Alex
2. The vertex is at -1, because you are shifting the graph of l 2x - 3 l down the y-axis by 1 step.

With f(x) = l2x -3l - 1, you have two transformations. Firstly, you reflect (2x - 3) in the x-axis for all points below the x-axis ( y negative ) - so your vertex would normally be at (3/2 , 0) and then you shift the whole graph down 1 step. So your new vertex is (3/2 , -1).

Hope that helps.
3. Cheers but also how do you work out the points where it crosses the graph

(2,0)
(1,0)
(0,2)

Thanks
4. (Original post by Alex H)
Cheers but also how do you work out the points where it crosses the graph

(2,0)
(1,0)
(0,2)

Thanks
i've attached the graphs of y=2x-3 and y=|2x-3|
from the graph, you can see that when x=0, the black line crosses at (0,-3), therefore the modulus of that graph crosses at (0,3). So when the graph is moved down one unit, it will corss the y-axis at (0,2)

for the other points, rearrange the equation to get:

y-1 = |2x-3|
square both sides: y^2 - 2y + 1 = 4x^2 - 12x + 9
y=0, 4x^2 - 12x + 8 = 0
x^2 - 3x +2 = 0
(x-2)(x-1) = 0
x=2 or x=1.....therefore the points are (2,0) and (1,0)
Attached Images

5. The graph crosses the X axis when y = 0

so solve the positive and negative of the modulus

2x-4 = 0
which gives x=2 and subsequently, y=0 =>. (2,0)

Then solve the minus of the modulus i.e. -(2x-3) - 1
which gives -2x + 3 -1 = -2x + 2
So solve -2x+2 = 0, giving x=1 and y=o, => (1,0)
6. Thanks people!

Does anybody know where I can get a couple more of those questions to practice?

Alex

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