True or False? 2^3 < 3^2
2 answers:
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Answer:
min at (3, 0 )
Step-by-step explanation:
given a quadratic in standard form
y = ax² + bx + c ( a ≠ 0 )
then the x- coordinate of the vertex is
= -
y = (x - 3)² = x² - 6x + 9 ← in standard form
with a = 1 and b = - 6 , then
= -
= 3
substitute x = 3 into the equation for corresponding value of y
y = (3 - 3)² = 0² = 0
vertex = (3, 0 )
• if a > 0 then vertex is minimum
• if a < 0 then vertex is maximum
here a = 1 > 0 then (3, 0 ) is a minimum
Answer:
(-√(6-√26) < x < √(6-√26)) ∪ (x < -√(6 +√26)) ∪ (√(6 +√26) < x)
Step-by-step explanation:
Using x^2 = z, the equation can be rewritten as ...
z^2 -12z +10 > 0
(z -6)^2 -26 > 0
|z -6| > √26
This resolves to two equations.
This one ...
x^2 -6 < -√26 . . . . substitute x^2 for z
|x| < √(6-√26) . . . . add 6, take the square root; use √a^2 = |a|
-√(6-√26) < x < √(6-√26)
__
and this one ...
x^2 -6 > √26
|x| > √(6 +√26)
x < -√(6 +√26) ∪ √(6 +√26) < x
