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Troyanec [42]
3 years ago
6

F(x)=-x^2+6x+14 vertex

Mathematics
2 answers:
ira [324]3 years ago
8 0
4 because the demented to d dj did do. Dkdndmd hdjdjd
sasho [114]3 years ago
4 0

Answer:

Vertex: (3, 23)

Step-by-step explanation:

a = -1   b = 6   c = 14

x vertex = \frac{-b}{2a}

\frac{-6}{-2}=3

f(x)-x^{2} +6x+14\\f(3) = -(3)^{2}+6(3) + 14 \\f(3)=-9+18+14\\f(3)=23

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<em>t ≥ -12 </em>

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Step-by-step explanation:

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3 years ago
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2 years ago
Show that the equation x^4/2021 − 2021x^2 − x − 3 = 0 has at least two real roots.
andreev551 [17]

The roots of an equation are simply the x-intercepts of the equation.

See below for the proof that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} has at least two real roots

The equation is given as: \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}

There are several ways to show that an equation has real roots, one of these ways is by using graphs.

See attachment for the graph of \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}

Next, we count the x-intercepts of the graph (i.e. the points where the equation crosses the x-axis)

From the attached graph, we can see that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} crosses the x-axis at approximately <em>-2000 and 2000 </em>between the domain -2500 and 2500

This means that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} has at least two real roots

Read more about roots of an equation at:

brainly.com/question/12912962

6 0
3 years ago
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