Thread: Quadratic Function + linear function

Write the following into the form of


and find the vertex, the symmetric axis and y-intercept

a)





Vertex = (-1,-2), Symmetric, x=-1, y-intercept, when x=0, y=-1

Am i doing the correct method to calculate ??


b)





Vertex = (2,3), symmetric x=2, y-intercept=-1

c)





Vertex (-1,-6), symmetrix =-1, y-intercept=-6

am i correct ?

Originally Posted by gilagila

Write the following into the form of


and find the vertex, the symmetric axis and y-intercept

a)







Vertex = (1,-2), Symmetric, x=1, y-intercept, when x=0, y=-1

Am i doing the correct method to calculate ??


b)




then what is the next step ? the answer is

same goes to

c)




[tex]f(x)=(x+1)^2-2

answer is

You can't factorise and then get rid of the other factors, and then say the functions are equal.

For the second, you should be writing



See what you can do for the third.

Originally Posted by Prove It

You can't factorise and then get rid of the other factors, and then say the functions are equal.

For the second, you should be writing



See what you can do for the third.

oops, just notice that my completing in square is wrong just now, able to correct, get the same answer as you.

how about the no.3, i make some changes, is it correct ?

Originally Posted by gilagila

Write the following into the form of


and find the vertex, the symmetric axis and y-intercept

a)







Vertex = (-1,-2), Symmetric, x=-1, y-intercept, when x=0, y=-1. Q:Note this correction.

b)
Q:Your working is incorrect.
Starting again:




Now solve for the vertices, symmetry etc.

same goes to

c)

Q:Again, incorrect. - the only time you can divide through by a common factor is when identifying roots.

Note the corrections.

Originally Posted by Quacky

Note the corrections.

oops, never notice for Q3 the k also need x3, ya..finally resolved all the question.

TQ