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y=-x^2-4x-8Identify the vertex, the axis of symmetry, the maximum or minimum value, and the range of the parabola.

Sagot :

Here we have the following parabola:

[tex]y=-x^2-4x-8[/tex]

To find the vertex, we could use the following formula:

[tex]V(x,y)=V(\frac{-b}{2a},\frac{-b^2}{4a}+c)[/tex]

Where a, b and c are the coefficients of the quadratic function:

[tex]y=ax^2+bx+c[/tex]

As you can see, in this problem a = -1 , b = -4 and c = -8. Thus,

[tex]V(x,y)=V(\frac{-(-4)}{2(-1)},\frac{-(-4)^2}{4(-1)}-8)[/tex]

This is:

[tex]V(-2,-4)[/tex]

Then, the vertex of the parabola is (-2,-4)

The axis of symmetry of the parabola is the line x=-2. Since the vertex is situated at the coordinates (-2,-4), that means that the parabola is symmetrical around this line.

The vertex is maximum point of the parabola.

The range, is defined as all the values that the y-axis could take. If we notice, that is:

[tex](-\infty,-4\rbrack[/tex]

I'm going to upload a picture of the parabola:

View image EdynnK27220
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