where are the zeros of this parabola?
x^2 + 6 x + 18 = 0
x = [-6 +/- sqrt (36-72)]/2
= -3 +/- 3
vertex is at x = -3
g has it at at x = 0
how high is it? roots are not real so above x axis
y = (-3)^2 -18 +18
= 9
so to be even
g(x) = x^2 + k
when x = 0, y must be 9
so
g(x) = x^2 + 9
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move that left 3
x' = x+3
x^2 + 6 x + 9
move that up 9
x^2 + 6 x + 18
The function f(x)=x2+6x+18 can be obtained from an even function g by shifting its graph horizontally and vertically. That even function is g(x)= ?
5 answers
g(x) = x^2
extra credit: shifted by how much in each direction?
extra credit: shifted by how much in each direction?
left 3, up 9... but the g(x) function is wrong or i might have a glitch in my online assignment
well g = x^2 + anything will work. Steve used zero
totally lost