Answer:
Step-by-step explanation:
A
The vertex form: f(x) = a(x-h)^2 + k
f(x) = x^2 + 16x + 60 = (x^2 + 16x) + 60
We want to get a perfect square in the brackets, so we solve for our b^2 coefficient.
b^2 = (16/2)^2 = 64
f(x) = (x^2 + 16x + 64 - 64) + 60. Note we subtracted 60 right away to end up with an equivalent expression and not some other function.
f(x) = (x+8)^2 - 4, as you can see it matches the general vertex form.
The vertex form shows when the profit is minimal. The point (h, k) or f(h).
B. The x-intercepts or when the function is equal to 0, or the profit is 0 in the context of the problem.
f(x) = x^2 + 16x + 60 set = 0
x^2 + 16x + 60 = 0
[tex]x_{12} = \frac{-16 \pm \sqrt{256 - 4(1)(60)}}{2} = \frac{-16 \pm \sqrt{16}}{2} = \frac{-16 \pm 4}{2} = -8 \pm 2[/tex]
Answer:
the vertex is either the maximum or minimum value
since the leading coefinet is negative (the number in front of the x² term), the parabola opens down and is a maximum
so
A.
a hack version is to use the -b/(2a) form
if you have f(x)=ax²+bx+c, then the x value of the vertex is -b/(2a)
so
given
f(x)=-1x²+16x-60
the x value of the vertex is -16/(2*-1)=-16/-2=8
the y value is f(8)=-1(8)²+16(8)-60=
-1(64)+128-60=
4
the vertex is (8,4)
so you selll 8 candels to make the max profit which is $4
B.
x intercepts are where the line crosses the x axis or where f(x)=0
solve
0=-x²+16x-60
0=-1(x²-16x+60)
factor
what 2 numbers multiply to get 60 and add to get -16
-6 and -10
0=-1(x-6)(x-10)
set each factor to 0
0=x-6
x=6
0=x-10
10=x
x intercepts are at x=6 and 10
that is where you make 0 profit
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Thank you
sincerely caitlin
