* = power of
(please answer asap, my exams tomorrow!) even if you have part of one question thats great too
1. The equation is y=(x-3)*2 - 16
I found ..
the zeros: (x+3) (x-6)
axis of symmetry: -1.5
and vertex: -1.5, 20.25)
how do you find 5 ordered pairs? and what info do i use to graph the relation?
2. What is the equation of the new parabola if the graph of y=x*2 is translated 3 units to the left?
3. If the parabola shown only gives me the vertex is (4,0) how do you find the equation? (points up)
(please answer asap, my exams tomorrow!) even if you have part of one question thats great too
1. The equation is y=(x-3)*2 - 16
I found ..
the zeros: (x+3) (x-6)
axis of symmetry: -1.5
and vertex: -1.5, 20.25)
how do you find 5 ordered pairs? and what info do i use to graph the relation?
2. What is the equation of the new parabola if the graph of y=x*2 is translated 3 units to the left?
3. If the parabola shown only gives me the vertex is (4,0) how do you find the equation? (points up)
-
you messed up #1
y = (x - 3)² - 16 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ this is a difference of squares, so
y = [(x - 3) + 4] [(x - 3) - 4] ∙ ∙ ∙and then simplify
y = (x + 1)(x - 7)
so zeros at -1 and 7, axis of symmetry x = (-1+7)/2 = 6/2 = 3
zeros are axis of symmetry ±4, so vertex is 4² down at (3, -16)
5 pairs? go 1 and 2 left and right of vertex, chanting 1 over, 1² up, 2 over, 2² up,
which gives you (1,-12), (2,-15), (3,-16), (4,-15), (5,-12)
by the way, standard notation for exponents is ^, so 3^2 = 9.
y = x^2 translated 3 left is y = (x + 3)^2
think of all parabolas as having 2 zeros, so if there's just 1, the 2 are the same,
which gives you y = (x - 4)²
depending on the vertical stretch you might make it y = a(x - 4)²
y = (x - 3)² - 16 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ this is a difference of squares, so
y = [(x - 3) + 4] [(x - 3) - 4] ∙ ∙ ∙and then simplify
y = (x + 1)(x - 7)
so zeros at -1 and 7, axis of symmetry x = (-1+7)/2 = 6/2 = 3
zeros are axis of symmetry ±4, so vertex is 4² down at (3, -16)
5 pairs? go 1 and 2 left and right of vertex, chanting 1 over, 1² up, 2 over, 2² up,
which gives you (1,-12), (2,-15), (3,-16), (4,-15), (5,-12)
by the way, standard notation for exponents is ^, so 3^2 = 9.
y = x^2 translated 3 left is y = (x + 3)^2
think of all parabolas as having 2 zeros, so if there's just 1, the 2 are the same,
which gives you y = (x - 4)²
depending on the vertical stretch you might make it y = a(x - 4)²