If Lylah completes the square for f(x)=x^2-12x+7 in order to find the minimum, she must write f(x) in the general form f(x)=(x-a)^2+b. What is the value of a for f(x)?

x^2-12x+7=0 move constant to other side

x^2-12x=-7 add the square of half the linear coefficient to both sides...(12/2)^2=36

x^2-12x+36=29 now the left side is a perfect square...

(x-6)^2=29 so if

f(x)=(x-a)^2+b then

f(x)=(x-6)^2-29

So a=6

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ApusApus ApusApus · Answer 2 · 2019-05-18T04:44:32+02:00

Answer:

[tex]a=6[/tex].

Step-by-step explanation:

We have been given that Lylah completes the square for [tex]f(x)=x^2-12x+7[/tex]. In order to find the minimum she must write f(x) in the general form [tex]f(x)=(x-a)^2+b[/tex].

We know that vertex form of a parabola is in format [tex]f(x)=(x-h)^2+k[/tex], where, (h,k) is the vertex of parabola.

To complete the square for given equation, we will set our given equation equals 0.

[tex]x^2-12x+7=0[/tex]

[tex]x^2-12x+7-7=0-7[/tex]

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

Now, we will add [tex](\frac{b}{2})^2[/tex] to both sides of our given equation.

[tex](\frac{12}{2})^2=(6)^2=36[/tex]

[tex]x^2-12x+36=-7+36[/tex]

[tex]x^2-12x+36=29[/tex]

[tex]x^2-12x+6^2=29[/tex]

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

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

[tex](x-6)^2-29=0[/tex]

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

Upon comparing our equation by vertex form of parabola, we can see that the vertex of parabola is at point [tex](6,-29)[/tex]. Therefore, the value of 'a' for f(x) is 6.