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#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?


  • Precalc -

    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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