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Which equation represents a parabola that opens upward, has a minimum at x = 3, and has a line of symmetry at x = 3?

[tex]A. y = x^2 - 3x + 6\\\\B. y = x^2 + 8x + 19\\\\C. y = x^2 + 6x + 5\\\\D. y = x^2 - 6x + 13[/tex]

Answer :

kudzordzifrancis

Answer:

[tex]y=x^2-6x+13[/tex]

Step-by-step explanation:

All the parabolas open upwards because they all have an 'a' value of a=1 which is positive.

The axis of symmetry of a parabola is calculated using the formula;

[tex]x=-\frac{b}{2a}[/tex]

We have x=3 and a=1.

We substitute the values to get;

[tex]3=-\frac{b}{2(1)}[/tex]

[tex]-2\times 3=-\frac{b}{2}\times -2[/tex]

[tex]b=-6[/tex]

Therefore the equation is of the form;

[tex]y=x^2-6x+c[/tex]

Looking at the given options, the required equation is

[tex]y=x^2-6x+13[/tex]

Answer:

D. y = x^2 -6x + 13

Step-by-step explanation:

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