If the y = ( x - 6)( x + 12) is graphed in the xy – plane, what is the x-coordinate of the parabola’s vertex?

Please with steps and formula thx!

The zeros of the function are found easily from the factors. They are 6 and -12. The x-coordinate is halfway between them, at (6 + (-12))/2 = -6/2 = -3.

If you want a formula, consider a quadratic with roots (zeros) p and q. Then it factors as

... y = (x -p)(x -q)

The line of symmetry through the vertex is also the line of symmetry between the roots, so has x-coordinate:

... x = (p+q)/2

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In your problem, you have p=6, q=-12, so x = (p+q)/2 = -6/2 = -3 is the line of symmetry and the x-coordinate of the vertex.

MrRoyal MrRoyal · Answer 2 · 2021-10-15T12:25:07+02:00

The vertex of a parabola is the minimum or the maximum point on the parabola

The x-coordinate of the vertex is -3

The equation is given as:

[tex]y = ( x - 6)( x + 12)[/tex]

Open brackets

[tex]y = x^2 - 6x +12x - 72[/tex]

[tex]y = x^2 + 6x - 72[/tex]

Differentiate

[tex]y' = 2x + 6[/tex]

Set to 0

[tex]2x + 6 = 0[/tex]

Collect like terms

[tex]2x =- 6[/tex]

Divide both sides by 2

[tex]x =- 3[/tex]

Hence, the x-coordinate of the vertex is -3

If the y = ( x - 6)( x + 12) is graphed in the xy – plane, what is the x-coordinate of the parabola’s vertex? Please with steps and formula thx!

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If the y = ( x - 6)( x + 12) is graphed in the xy – plane, what is the x-coordinate of the parabola’s vertex?

Please with steps and formula thx!