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What is the line of symmetry for the parabola whose equation is y=x²-12x + 7?
Ox=12
Ox=6
Ox=-6

Sagot :

Fieryanswererftidn

Answer:

x = 6

Explanation:

completing square:

[tex]y=x^2-12x+7[/tex]

[tex]y=(x^2-12x)+7[/tex]

[tex]y=(x-6)^2+7-(-6)^2[/tex]

[tex]y=(x-6)^2+7-36[/tex]

[tex]y=(x-6)^2-29[/tex]

Comparing with quadratic equation [tex]y=ax^2 + bx+c[/tex], in vertex form where [tex]y = a(x-h)^2+k[/tex]. In this x - h = 0, x = h defines the symmetry of equation.

So here the symmetry for parabola:

x - 6 = 0

x = 6

Semsee45idn

Answer:

x = 6

Step-by-step explanation:

The axis of symmetry of a parabola is the x-value of its vertex.

For a quadratic function in the form  [tex]y=ax^2+bx+c[/tex], the x-value of the vertex is:

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

Given function:

[tex]y=x^2-12x+7[/tex]

Therefore:

[tex]a=1, \quad b=-12, \quad c=7[/tex]

So the axis of symmetry of the given quadratic function is:

[tex]\implies x=-\dfrac{b}{2a}=-\dfrac{-12}{2(1)}=\dfrac{12}{2}=6[/tex]

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