Question

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)?

Answer 1

Answer:

$a=6$.

Step-by-step explanation:

We have been given that 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 $f(x)=(x-a)^2+b$.

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

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

$x^2-12x+7=0$

$x^2-12x+7-7=0-7$

$x^2-12x=-7$

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

$(\frac{12}{2})^2=(6)^2=36$

$x^2-12x+36=-7+36$

$x^2-12x+36=29$

$x^2-12x+6^2=29$

$(x-6)^2=29$

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

$(x-6)^2-29=0$

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

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

Answer 2

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