What is the vertex of the function?
1 answer:
Answer:
(4, -25)
Step-by-step explanation:
One way to answer this is to put the equation into vertex form.
f(x)= x^2 -8x -9 . . . . . given
Add and subtract the square of half the x-coefficient:
f(x) = x^2 -8x +(-8/2)^2 -9 -(-8/2)^2
f(x) = x^2 -8x +16 -25
f(x) = (x -4)^2 -25
Comparing this to the vertex form of a quadratic:
f(x) = a(x -h)^2 +k
we find that (h, k) = (4, -25). This is the vertex.
_____
<em>Alternate solution</em>
The line of symmetry for ...
f(x) = ax^2 +bx +c
is given by x = -b/(2a). For your given quadratic that line is ...
x = -(-8)/(2(1)) = 4
Evaluating f(4) gives the y-coordinate:
f(4) = 4^2 -8·4 -9 = -25
The vertex is (x, y) = (4, -25).
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Answer:
The function h (x) = StartFraction 11 Over x minus 4 EndFraction
has the domain of h (x) = ( - ∞, 4) U (4, ∞ ) which is same as the
domain of (m ∘ n)(x) = ( - ∞, 4) U (4, ∞ ).
Step-by-step explanation:
m (x) = StartFraction x + 5 Over x minus 1 and n(x) = x – 3
As m (x) could be written as:
n(x) = x – 3
(m ∘ n)(x) = m(n(x)) = m(x – 3)
=
=
In order to find the domain of (m ∘ n)(x) = , we need to
make sure that denominator can not be zero,
So,
x - 4 = 0
x = 4
So, our domain can be anything except for 4.
Hence, Domain = D = ( - ∞, 4) U (4, ∞ )
Now, compare the domain of (m ∘ n)(x) i.e D = ( - ∞, 4) U (4, ∞ ) with all the options:
Option A: h (x) = x + 5 / 11
Option A has the Domain of h (x) = Set of all real numbers
So, Option A is false.
Option B: h (x) = 11 / x - 1
x - 1 = 0
x = 1
Option B has the domain of h (x) = ( - ∞, 1) U (1, ∞ )
So, Option B is false.
Option C: h (x) = 11 / x - 4
x - 4 = 0
x = 4
Domain C has the domain of h (x) = ( - ∞, 4) U (4, ∞ )
So, Option C is true.
Option D: h (x) = 11 / x - 3
x - 3 = 0
x = 3
Option D has the domain of h (x) = ( - ∞, 3) U (3, ∞ )
So, option D is also false.
Hence, only option C is true as the option C i.e. h (x) = StartFraction 11 Over x minus 4 EndFraction has the domain of h (x) = ( - ∞, 4) U (4, ∞ ) which is same as the domain of (m ∘ n)(x) = ( - ∞, 4) U (4, ∞ ).
keywords: Domain, composite function
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