The graph of a function is a parabola that has a minimum at the point (-3,9). Which equation could represent the function?
1 answer:
Well there are no choices listed so it's going to be hard to say which; we'll write an expression for all of them and then give a few instances.
A parabola with a minimum forms a CUP, concave-up positive, meaning the coefficient a on x² must be positive, a>0.
The general form with vertex (p,q) is
y = a(x-p)² + q
So for us, all our parabolas are of the form
y = a(x- -3)² + 9
y = a(x² + 6x + 9) + 9
y = ax² + 6ax + 9(a+1)
That's the general form for a parabola with vertex (-3,9); a>0 assure the parabola has a minimum at the vertex.
Some instances:
a=1 gives
Answer: y = x²+6x+18
a=4 gives
y = 4x² + 24x + 45
Other positive <em>a</em>s give other possible answers; without the choices it's impossible to know which one they're seeking.
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Answer: t = - 5 ∈ = ( -∞ , -4 )
Step-by-step explanation:
The standard form of O.D.E is written as :
+
Equation given :
+
=
,
The first thing to do is to write the O.D.E in standard form , that is we will divide through by , so we have
With this , we can see that and
are both continuous in the same domain. Therefore , the intervals are :
= ( -∞ , -4 )
= ( - 4 , 4 )
= ( 4 , -∞ )
recall that y(−5) = 1 , then t = -5
This means that :
t = - 5 ∈ = ( -∞ , -4 )