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Karo-lina-s [1.5K]
2 years ago
8

3.

Mathematics
1 answer:
Mekhanik [1.2K]2 years ago
5 0

Hello Papi :D

I will solve the problem by applying the perfect square trinomial. In this way we obtain the canonical form. Another way would be to derive the function, but I don't know if you're familiar with it.

f(x)=-{x}^{2}-3x+6

First: let us take out the common factor: -1, since we remember that the canonical form is characterized as follows:

\boxed{f(x)=a{(x-h)}^{2}+k

Then, it remains:

f(x)=-1({x}^{2}+3x-6)

Then: the coefficient of the variable x We divided it between 2, And we square it (they will be one positive and one negative). In our case:

\dfrac{3}{2}\rightarrow {\dfrac{3}{2}}^{2}

\frac{9}{4}[\tex]We apply it to the function:[tex]f(x)=-1({x}^{2}+3x-6+ \boldsymbol{\dfrac{9}{4}}- \boldsymbol{\dfrac{9}{4}})

Let's accommodate terms to make it easier:

f(x)=-1(\underline{{x}^{2}+3x+\frac{9}{4}}-6-\frac{9}{4}

-6 Can be written as -\frac{24}{4}:

f(x)=-1(\underline{{x}^{2}+3x+\frac{9}{4}}-\frac{24}{4}-\frac{9}{4}

f(x)=-1(\underline{{x}^{2}+3x+\frac{9}{4}}-\frac{33}{4}

Now, what is underlined is our perfect square trinomial, let us recall its form:

\boxed{{a}^{2}+2ab+{b}^{2}}

Applying the same principle we are left:

f(x)=-1[{(x+\frac{3}{2})}^{2}-\frac{33}{4})

Applying distributive property we get:

\boxed{\boxed{\boxed{f(x)=-{(x+\frac{3}{2})}^{2}+\frac{33}{4}}}}

Therefore it will have its vertex in: (-1.5,\:8.25)

The axis of symmetry is a straight line that makes the function to be projected being 2, for this you need some reference point, for the parabola you need the coordinate in x of the vertex.

For which the axis of symmetry is -1.5.

I love you so much !

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