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)
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)
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?
- 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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