Posted by **Rebekah** on Tuesday, September 25, 2012 at 9:23pm.

#1: f(x) = x^2-4x

find vrtex, axis ofsym, y-int, x-int if any, domain, range, where increasing and decreasing.

I got.... vertex: 2,4

axis of sym: x=2

D= (-infinity, infnity)

range= [4,infinity)

im stuck of the intercepts because when i plug i 0 for x i get yint of 0 but the graph doesnt show that.

Am i doing the completingthe square wrong

I got (x^2-2)^2 + 4

#2 Same requirements as the first question

f(x)= x^2 + 6x +9

I got... vertex: (-3,18)

axis of sym: x=-3

Domain= (-infinity, infinity)

Range= [18, infinity)

y-int (0,9)

x-int idk???

After completing the square I got:

(x+3)^2 + 18

Also a question, how do you know if there is an x or y int without drawing the graph because my graph des not show a xint but the equation does?

Thanks!

- Precalc -
**Scott**, Wednesday, September 26, 2012 at 1:14pm
you seem to be finding the axis of symmetry okay

when you plugged in to find the vertex in #1, there is an error (should be 2,-4)

__ this also affects the range

completing the square should be:

f(x) = (x - 2)^2 - 4

(x - 2)^2 = x^2 - 4x + 4

so you need to subtract out the extra 4 that you add when using the binomial-square

looks like a similar error on the vertex for #2 __ also in the range

#2 is already a perfect square, so completing the square went amiss

finding intercepts without a picture means individually (not at the same time) substituting zero for x and y, and solving for the other value

for quadratics in x, there will ALWAYS be a y-intercept

depending on the values of a, b, and c; there may be two, one , or no x-intercepts

the discriminant (b^2 - 4ac) shows the nature of the roots (x-intercepts)

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